how to find the quadratic equation from a graph

Its height, in meters above sea level, as a function of time, in seconds, is given by In this text, we will determine at least five points as a means to produce an acceptable sketch. This video explains how to determine the equation of a quadratic function from a graph. The$y$-coordinate of the vertexis $y = f(h)=k$. k=4. Answer. We know that Solve the quadratic equation $f(x)=0$ to find the $x$-intercepts. There are several actions that could trigger this block including submitting a certain word or phrase, a SQL command or malformed data. x times x minus two, and in these second two I can factor out. product of three and two? c So we could figure out what is x when, or what is y when x is equal ( Many quadratic equations cannot be solved by factoring. What appears to be the effect of adding a constant? Draw the graph of $y = x^2 -x - 4 $ and use it to find the roots of the equation to 1 decimal place. Solving quadratic equations by factoring. Created by Sal Khan. TblStart=6 $x$-intercept: none, Domain:$ (-\infty, \infty) $ 4x+13=0 is height in feet. 2 2a $\begin{array}{l}{y=a(x-h)^{2}+k} \\ \color{Cerulean}{\qquad\qquad\:\downarrow\quad\:\:\:\downarrow}\\ {y=2(x-1)^{2} \:+6}\end{array}$. (2,0). If the leading coefficient a is positive, then the parabola opens upward and there will be a minimum y-value. so the graph becomes narrower. In this case, solve using the quadratic formula with a = 1, b = 2, and c = 1. Direct link to A/V's post Yes that is also another , Posted 2 years ago. Some involve geometric approaches. To find the x-intercepts, set y = 0. +4. y= What is the product? The domain of a quadratic function is all real numbers. ]. x+3 =\pm \sqrt{ \dfrac{5}{2}}\\ Given a quadratic equation of the form $y=ax^{2}+bx+c$, find the y-intercept by setting x=0 and solving. We must then add $3$ to not change the value of the function. Step 4: Determine extra points so that we have at least five points to plot. ) In physics, for example, they are used to model the trajectory of masses falling with the acceleration due to gravity. ), Accessibility StatementFor more information contact us atinfo@libretexts.org. In this case, the quadratic can be factored easily, providing the simplest method for solution. The vertex is (1, 3). When in this form, the vertex is, The height in feet reached by a baseball tossed upward at a speed of 48 feet/second from the ground is given by the function $h(t)=16t^{2}+48t$, where, The height of a projectile launched straight up from a mound is given by the function $h(t)=16t^{2}+96t+4$, where, The profit in dollars generated by producing and selling, The revenue in dollars generated from selling a particular item is modeled by the formula $R(x)=100x0.0025x^{2}$, where, The average number of hits to a radio station website is modeled by the formula $f(x)=450t^{2}3,600t+8,000$, where, The value in dollars of a new car is modeled by the formula $V(t)=125t^{2}3,000t+22,000$, where, The daily production costs in dollars of a textile manufacturing company producing custom uniforms is modeled by the formula $C(x)=0.02x^{2}20x+10,000$, where. One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, Direct link to JK223's post tow steps were involved. x a<0, t Once the equation is in this form, we can easily determine the vertex. Write down your plan for graphing a parabola on an exam. \text{Since }a \text{ ispositive, the parabola opens upward. } a>0, f( The quadratic equation in its standard form is ax 2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. 10x+26=0 Graph. We can see the maximum and minimum values in Figure 9. $\begin{array}{l}{y=\:a\:(\:x-h)^{2}+k} \\ \color{Cerulean}{\qquad\qquad\quad\downarrow\quad\:\:\downarrow} \\ {y=-4(x-3)^{2}+1}\end{array}$. f(x)= A quadratic function is a polynomial function of degree two. Questions Tips & Thanks Want to join the conversation? Because there are no real solutions, there are no x-intercepts. f(x)k. So our x value is 1/3. x ,0 x The ball reaches a maximum height of 140 feet. 2x+4=0 A Quadratic Function is any function defined by a polynomial whose greatest exponent is two. \nonumber\]. x=2 x For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Example $\PageIndex{6}$: Finding the $y$- and $x$-intercepts of a General Form Quadratic. 2 a Solve f(x)k; 2,4 the graph shifts to the left. 4x+2 (1,1) x 4x+2. 2 Expand and simplify to write in general form. If you are redistributing all or part of this book in a print format, Graph the vertex, intercepts, and the point symmetric to the $y$-intercept. ). x- To this: f (x) = a (x-h)2 + k Where: h = b/2a k = f ( h ) In other words, calculate h (= b/2a), then find k by calculating the whole equation for x=h But Why? x- opens down. Find the domain and range of x +2 If you need a review on solving quadratic equations, feel free to go to Tutorial 17: Quadratic Equations. 2 And so now we can Substituting these values into the formula produces, \[\begin{align*} x&=\dfrac{1{\pm}\sqrt{1^241(2)}}{21} \\ &=\dfrac{1{\pm}\sqrt{18}}{2} = \dfrac{1{\pm}\sqrt{7}}{2}=\dfrac{1{\pm}i\sqrt{7}}{2} \nonumber \end{align*}\]. 7 . Draw and complete a table of values to find coordinates of points on the graph. What two algebraic methods can be used to find the horizontal intercepts of a quadratic function? Q=79,000. Q=2,500p+159,000 Now we are ready to write an equation for the area the fence encloses. And what's interesting about The standard form of a quadratic function presents the function in the form. The y-intercept is the point where the graph intersects the y-axis. How many trees should she plant per acre to maximize her harvest? a=3,h=2, x+2 \end{array} \), $ \begin{array}{llc} 2 The general form of a quadratic equation is ax 2 + bx + c = 0, where a, b and c are real numbers, also called " numeric coefficients" and a 0. h( The y-intercept is the point at which the parabola crosses the y-axis. Find the domain and range of x 3 \(\begin{aligned} x &=\frac{-b}{2 a} \\ &=\frac{-(\color{OliveGreen}{-2}\color{black}{)}}{2(\color{OliveGreen}{1}\color{black}{)}} \\ &=\frac{2}{2} \\ &=1 \end{aligned}$. y- 61 2 a (x,y) The horizontal coordinate of the vertex will be at, The vertical coordinate of the vertex will be at. Given a quadratic equation of the form $y=ax^{2}+bx+c$, x is the independent variable and y is the dependent variable. We already typed that in. and has the shape of How to: Graph a quadratic function in the form $f(x)=a(x-h)^2+k$, Example $\PageIndex{10}$: How to Graph a Vertex Form Quadratic Using Properties, Graph the function $f(x)=2(x+1)^{2}+3$ by using its properties. 6 Once you've done that, refresh this page to start using Wolfram|Alpha. It seems to be the point two comma one. &=-2(3)^2+5\\ The vertex is the point that defines the minimum or maximum of the graph. $\begin{aligned} y &=2 x^{2}+4 x+5 \\ &=2(\color{OliveGreen}{-1}\color{black}{)}^{2}+4(\color{OliveGreen}{-1}\color{black}{)}+5 \\ &=2-4+5 \\ &=3 \end{aligned}$. two times three x minus one. TABLE. In standard form, the algebraic model for this graph is h( For the following exercises, use the table of values that represent points on the graph of a quadratic function. b x= x f( h( (h,k) 2 $\begin{array}{c}{y=a(x-h)^{2}\:+\:\:k} \\ \color{Cerulean}{\qquad\quad\quad\:\:\downarrow\qquad\downarrow} \\ {y=(x-(-2))^{2}+5}\end{array}$. y- f(x) 2 The range is bounded by the, An alternate approach to finding the vertex is to rewrite the quadratic equation in the form $y=a(xh)^{2}+k$. opens up. (x+2) Rewrite in $y=a(xh)^{2}+k$ form and determine the vertex: $y=x^{2}+4x+9$. Yes that is also another fast way to do it. 2 (1,4) 3 In addition, find the x-intercepts if they exist. x The range is $f(x){\leq}\frac{61}{20}$, or $\left(\infty,\frac{61}{20}\right]$. Step 4: Find the $y$-intercept. 2 Rewrite in $y=a(x-h)^{2}+k$ form and determine the vertex: $y=-2(x+3)^{2}+21$; vertex: $(-3, 21)$, Exercise $\PageIndex{4}$ the graph of quadratic equations. Example $\PageIndex{5}$: Findthe Domain and Range of a Quadratic Function. Question: Find the equation of the quadratic function g whose graph is shown below. Therefore, the y-value of the vertex determines the maximum height. 2 Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function. +x+2=0. 2 +5x2. So far, we have only two points. To begin, we graph our first parabola by plotting points. ) intercepts at and Here you can get a visual of your quadratic function Worksheets x+1 = \pm \sqrt(-3/2) \\ ( standard form of a quadratic function For the equation 2 Find additional points if needed. ), 2 Range: $[1, \infty)$, There are two important forms of a quadratic function, Definitions: Forms of Quadratic Functions. (xh) h( k<0, The standard form and the general form are equivalent methods of describing the same function. x- . A quadratic equation in two variables, where a,b,and c are real numbers and a 0, is an equation of the form y = a x 2 + b x + c Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations. x We connect these $5$ points to sketch the parabola. (x+2) However, the range (the set of y-values) is bounded by the y-value of the vertex. Contains )= 2 Roots, solutions, graphs. We need one more point. intercepts can vary depending upon the location of the graph. h=2. +4x4. Wolfram|Alpha can apply the quadratic formula to solve equations coercible into the form . And then I can factor out a negative two. This is actually a 2 h(x)=.0001 x- +4x+3. and has the shape of Subtract the CTS constant (multiplied by the coefficient of$x^{2}$ if not 1). 1 Substitute y = 2 y = 2 in the equation to find the x x -coordinate of . Determine the vertex, axis of symmetry, zeros, and Tbl=2, Therefore, the minimum y-value of 2 occurs when x = 4, as illustrated below: A parabola, opening upward or downward (as opposed to sideways), defines a function and extends indefinitely to the right and left as indicated by the arrows. Use the Square Root Property. (h,k)=(2,0),(x,y)=(4,4) 2 Find the point symmetric to the$y$-intercept across the axis of symmetry. +6x+4 a>0, a<0, f(x)=2 +36=0 +10x+12 Discovered in ancient times, the quadratic formula has accumulated various derivations, proofs and intuitions explaining it over the years since its conception. \text{The result is the }y\text{ coordinate of the vertex.} The steps for graphing a parabola are outlined in the following example. Cloudflare Ray ID: 7d2a76f8f8dc1247 x This is a formula for the $x$-coordinate of the vertex. Here c = 5 and the y-intercept is (0, 5). Write the trinomial as a binomial square and combine constants outside the binomial square to arrive at the standard form of the function. f(x)=2 y=-2 (x+5)^2+4 y = 2(x + 5)2 + 4 This equation is in vertex form. t A quadratic is a polynomial of degree two. g(x)=13+ x we get three x squared minus seven x plus two. intercept of the parabola shown in Figure 3. Given a quadratic function, find the domain and range. the graph shifts toward the right and if + f( 6x+10=0 To do this, set x = 0 and solve for y. both of these equations. 1 For the $x$-intercepts, find all solutions of $f(x)=0$. x 2 The maximum value of the function is an area of 800 square feet, which occurs when How to: Rewrite $y=ax^2+bx+c$ into vertex form - complete the squaremethod. After identifying parameters $a$ and $b$,calculate $h=\dfrac{b}{2a}$. 2( Why is any parabola that opens upward or downward a function? The $x$-interceptsare the points where the graph crosses the $x$-axis. What is another name for the standard form of a quadratic function? ) Values of the parameters in the general form are $a=2$, $b=4$, and $c=-4$. Completing the square, factoring and graphing are some of many, and they have use casesbut because the quadratic formula is a generally fast and dependable means of solving quadratic equations . Or is it important to learn this way? \text{a.} $y=(x7)^{2}25$; vertex: $(7, 25)$, 3. a=3,h=2, x is in thousands of phones produced, and the revenue represented by thousands of dollars is h=2. However, since a parabola is curved, we should find more than two points. To write this in general polynomial form, we can expand the formula and simplify terms. . For the following exercises, use a calculator to find the answer. Factor from both terms: The factors of are and . divides the graph in half. !, Posted 9 months ago. (x+2) \text{a.} If $a$ is positive, the parabola has a minimum value of$k$ and the, If $a$ is negative, the parabola has a maximum value of$k$ and the. x- With a ticket price of $11, the average attendance has been 26,000. opens down. 1 So I'm going to get three Since it is quadratic, we start with the $f(x)=a(xh)^{2}+k$ form. The basic idea behind solving by graphing is that, since the (real-number) solutions to any equation (quadratic equations included) are the x-intercepts of that equation, we can look at the x-intercepts of the graph to find the solutions to the corresponding equation.However, there are difficulties with "solving" this way. 3. f(x)=3 We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, (h,k) Since. The axis of symmetry is the line $x=-1$. x 5 Questions Tips & Thanks Want to join the conversation? Determine whether the graph of each function is a parabola that opens upward or downward: When given a quadratic in standard form$f(x)=a(xh)^2+k$, the vertex and axis of symmetry is easily found once the parameters $h$ and $k$ have been identified. 3. $ \quad $ Since the $a$ is negative, the parabola will open downward. x Example $\PageIndex{13}$: Writing the Equation of a Quadratic Function from the Graph. x Write an equation for the quadratic function $g$ in Figure $\PageIndex{13}$ in standard (vertex) form and then rewrite the result intogeneral form. 2 One important feature of the graph is that it has an extreme point, called the vertex. +k. Contains Graph on the same set of axes the functions If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of substitute x minus one back in for y, and so we get 2 ). +5x2. x Solving a quadratic equationgiven in standard form, like in this example, is most efficiently accomplished byusing the Square Root Property, $ \begin{array}{c} Contains \(f(0)=(\color{red}{0}\color{black}{)}^{2}-6(\color{red}{0}\color{black}{)}+8$ f( So the $y$-intercept is at $(0,13)$. Find the vertex and the line of symmetry. The standard form is useful for determining how the graph is transformed from the graph of Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down. value is Therefore, the domain of any quadratic function is all real numbers. 2 Factor the coefficient of $x^{2}, -3$. 2 1. 2 +8x10, k( In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. 0 =-2(x+3)^2+5\\ The function, written in general form, is. b x and For the $x$-intercepts, find all solutions of $f(x)=0$. x The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given by the function a Does the shooter make the basket? A formula for the location of the vertex for a quadraticin general form can be found by equating the two forms for a quadratic. I'll scroll down a little 3 value is Find the minimum or maximum value of the quadratic function $f(x)=x^{2}+2 x-8$. f(x)= x Using the vertex to determine the shifts. ,0). 2 & \text{The axis of symmetry is the vertical line }x=h& \\ We know that any linear equation with two variables can be written in the form $y=mx+b$ and that its graph is a line. 7 h, , & \text{Identify the equation parameters}& a=3, b=-6, c=2 \\ And on the right-hand side (x+4) \\ A farmer finds that if she plants 75 trees per acre, each tree will yield 20 bushels of fruit. I see two intersection points. Accessibility StatementFor more information contact us atinfo@libretexts.org. \text{Substitute and simplify.} (1,2), For any parabola, we will find the vertex and y-intercept. There are certain key features that are important to recognize on a graph and to calculatefrom an equation. Thus the vertex, originally atat $(0,0)$, is located at the point $(h, k)$ in the graph of $f$. (h,k) )=2 x x & \text{Substitute.} Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. - [Instructor] We're x 2 h( | x | We know that the quadratic formula used to find the . f(x)= L. \\ &=4 \end{aligned}\). (x2) In Figure 5, x Begin by making room for the constant term that completes the square. f(x)= (1+ (1,6) If you don't know how, you can find instructions. In this case, the quadratic can be factored easily, providing the simplest method for solution. x The magnitude of x 's post at 4:45 where did he get , Posted 3 months ago. x x Since c = 18, the y-intercept is (0, 18). and In other words, we will need to solve the equation 0 = ax2 + bx + c for x. y = ax2 + bx + c 0 = ax2 + bx + c. Exercise $\PageIndex{5}$ the graph of quadratic equations, 1. x-intercepts: $(6, 0), (2, 0)$; y-intercept: $(0, 12)$, 3. x-intercepts: $(3, 0), (\frac{1}{2}, 0)$; y-intercept: $(0, 3)$, 5. x-intercepts: $(1, 0), (\frac{2}{5}, 0)$; y-intercept: $(0, 2)$, 7. x-intercepts: $(\frac{5}{2}, 0), (\frac{5}{2}, 0)$; y-intercept: $(0, 25)$, 9. x-intercepts: none; y-intercept: $(0, 1)$, Exercise $\PageIndex{6}$ the graph of quadratic equations. f(x)= 2 2 Often, the simplest way to solve " ax2 + bx + c = 0 " for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. Look $a$ in the equation $f(x)=x^{2}-6x+8$. 5x1 Q=2,500p+159,000 ( Therefore, the domain of any quadratic function is all real numbers. x- x- and You get three x is equal to one, or x is equal to 1/3. k<0, 1. Two more points: Given a quadratic function to represent the width of the garden and the length of the fence section parallel to the backyard fence. +6x+25=0 Now we could try to factor this. ) k>0, 2 x In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Use the value found for $h$ to find $k$. f(x)=2 Each method has its pros and cons. Point symmetric to $y$-intercept is $ (-2,5) $. ,f(x)=2 $y$-intercept at $(0, 13)$, No $x$-intercepts. y- In this example, one other point will suffice. (h,k)=(1,0),(x,y)=(0,1). a>0, 1+i It also reveals whether the parabola opens up or down. h( x 1 g(x)=13+ (1,2), is imposed in the definition of the quadratic function. 4ac 7 ,f( By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function. y- From the original equation, a = 2, b = 12, and c = 18. x- && \text{The vertex is } (1, -1)\\ 10x+26=0, x And so we are going to get y, and then all the rest of The axis of symmetry is the line x= b 2 a. Adding and subtracting the same value within an expression does not change it. In this example, choose the x-values {2, 1, 0, 1, 2, 3, 4} and calculate the corresponding y-values. f(x)k; 2 x 2 +5x2 intercepts (if any). 2 the equation for the axis of symmetry. 2 2 feet. f(x)=a Another way of transforming$f(x)=ax^{2}+bx+c$ into the form $f(x)=a(xh)^{2}+k$ is by completing the square. Rewrite the following functions in the $f(x)=a(xh)^{2}+k$ form by completing the square. 2 1i Furthermore, c = 1, so the y-intercept is (0, 1). x x= value of the vertex. 2 and this is one way to tackle any system of equations, 3,1 x find the The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b (b^2 - 4ac)) / (2a) Does any quadratic equation have two solutions? Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing. The x-intercepts are the points at which the parabola crosses the x-axis. 1 comment ( 20 votes) Upvote in this example, +27=0, x x The vertex is the turning point of the graph. x x- When the curve crosses the x-axis (y=0) you will have: two solutions or ONE solution (if it just touches) +x+2=0. How features of the parabola for a quadratic function can be obtained is summarized below. $f(x)=x^{2}-6x+8$ Direct link to Timo's post Please check out factorin, Posted 3 years ago. $ \begin{array}{llc} Vertex: \((\frac{3}{2}, 2)$; line of symmetry: $x= \frac{3}{2}$, 5. (x3) \text{b.} (h,k)=(3,2),(x,y)=(10,1), (h,k)=(0,1),(x,y)=(1,0) f(x)f( or Determine the quadratic function whose graph is shown. We reviewed their content and use your feedback to keep . so this is the y-intercept. x = 2 o r x = 5 A quadratic equation has two roots if its graph has two x-intercepts A quadratic equation has one root it its graph has one x-intercept A quadratic equation has no real solutions if its graph has no x-intercepts. \end{array} \). To determine three more, choose some x-values on either side of the line of symmetry, x = 1. We can see the maximum revenue on a graph of the quadratic function. Can I think of two numbers, a times b, that's equal to the The vertex can be found from an equation representing a quadratic function. Its height, in meters above ground, as a function of time, in seconds, is given by A ball is thrown in the air from the top of a building. f(x)= ). Direct link to David Smith's post At 5:20, why does solving, Posted 3 years ago. Now, these features will be used to sketch a graph. The area of a certain rectangular pen is given by the formula $A=14ww^{2}$, where. $y$-intercept: $(0,7)$ Since a = 4, we know that the parabola opens downward and there will be a maximum y-value. This approach will also be used when circles are studied. Rewritethe quadratic function $f(x)=2x^2+4x4$ into standard form. 2 Since the solutions are imaginary, there are no $x$-intercepts. x y- A rocket is launched in the air. This gives us the linear equation We begin by solving for when the output will be zero. x= H(t)=16 0,7 are real numbers and b Vertex It is often useful to find the maximum and/or minimum values of functions that model real-life applications. So $x=0$ and $f(0)=7$. x +36=0, x So we can see the parabola here in red, and we can see the line here in blue. ,f(x)= Determine whether $a$ is positive or negative. 2 (h,k)=(5,3),(x,y)=(2,9), (h,k)=(3,2),(x,y)=(10,1) The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. andf(x)= 2 and (2,1). a +bx+c The other method uses Complete the Square. (x3) R=xp. the point associated with a particular f( In Example 7, the quadratic was easily solved by factoring. \[\begin{align*} 0&=3x1 & 0&=x+2 \\ x&= \frac{1}{3} &\text{or} \;\;\;\;\;\;\;\; x&=2 \end{align*} \]. +4. where x equals two. 2 And can I think of those same two, a plus b, where it's going to be equal to negative seven? 6x 9: Solving Quadratic Equations and Graphing Parabolas, { "9.01:_Extracting_Square_Roots" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.02:_Completing_the_Square" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.03:_Quadratic_Formula" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.04:_Guidelines_for_Solving_Quadratic_Equations_and_Applications" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.05:_Graphing_Parabolas" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.06:_Introduction_to_Complex_Numbers_and_Complex_Solutions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.0E:_9.E:_Review_Exercises_and_Sample_Exam" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Real_Numbers_and_Their_Operations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Linear_Equations_and_Inequalities" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Graphing_Lines" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Solving_Linear_Systems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Polynomials_and_Their_Operations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Factoring_and_Solving_by_Factoring" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Rational_Expressions_and_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Radical_Expressions_and_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Solving_Quadratic_Equations_and_Graphing_Parabolas" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Appendix_-_Geometric_Figures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:anonymous", "licenseversion:30", "program:hidden", "cssprint:dense" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FAlgebra%2FBeginning_Algebra%2F09%253A_Solving_Quadratic_Equations_and_Graphing_Parabolas%2F9.05%253A_Graphing_Parabolas, $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}$ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$, 9.4: Guidelines for Solving Quadratic Equations and Applications, 9.6: Introduction to Complex Numbers and Complex Solutions, Finding the Vertex by Completing the Square, $(1-\sqrt{2}, 0)$ and $(1+\sqrt{2}, 0)$. 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