Chapter 9.1, Problem 70HP

To determine

To Find:

Using the axis of symmetry and one x-intercept, write an equation for the graph shown.

Answer to Problem 70HP

Hence, the required equation of the graph is

  $y=-x^2+6x+16$

Explanation of Solution

Given Information:

The given graph is :

  

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=-\frac{b}{2a}$

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
• Graph:

  

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

Calculation:

• The graph is a downward parabola. So, a < 0 .
• The vertex of graph is (3, 25).
• So, the axis of symmetry is the line x = 3.

Substitute the value of x in the equation of axis of symmetry and solving for b in terms of a :

  $\begin{array}{l}x=-\frac{b}{2a}=3\\ \Rightarrow -\frac{b}{2a}=3\\ \Rightarrow \frac{-b}{2a}\cdot (-2a)=3(-2a)\mathrm{...............}[multiply"-2a"each\_side]\\ \Rightarrow b=-6a\end{array}$

• The parabola has y-intercept at (0,16). So, we have c = 16.
• The parabola has x-intercepts at (-2,0) and (8,0).
• Now, the standard form of quadratic equation is $y=a{x}^2+bx+c$

Substituting the values of c = 16 , b = -6a and (x,y) = (8,0) in the above equation to solve for a :

  $\begin{array}{l}y=a{x}^2+bx+c\\ y=a{x}^2+(-6a)x+\mathrm{16...............}[c=16,b=-6a]\\ $ 0=a{(8)}^2-6a(8)+\mathrm{16................}[(x,y)=(8,0)]$0=64a-48a+16\\ 0=16a+\mathrm{16..............................}[simplify]\\ -16=16a+16-\mathrm{16..................}[subtract"16"each\_side]\\ 16a=-16\\ \frac{16a}{16}=\frac{-16}{16}\mathrm{...............................}[divide"16"each\_side]\\ a=-1\\ Now,\\ b=-6a=-6(-1)=6\end{array}$

• So, we have a = -1 , b = 6 , c = 16 .

Substituting these values in the standard form of quadratic equation:

  $\begin{array}{l}y=a{x}^2+bx+c\\ y=-x^2+6x+\mathrm{16.................}[a=-1,b=6,c=16]\end{array}$

• Hence, the required equation of the graph is

$y=-x^2+6x+16$