Chapter 1.2, Problem 8E

To determine

To find: The expressions for the given quadratic graphs.

Answer to Problem 8E

The expression of the graph f is $f\left(x\right)=2{\left(x-3\right)}^2$.

The expression of the graph g is $g\left(x\right)=-x^2-2.5x+1$.

Explanation of Solution

Consider the graph of the function f.

Here, the graph represents that an open upward parabola with vertex (3, 0). Moreover, it is observed that the point (4, 2) is lies on the parabola.

The general equation of the open upward parabola is, ${\left(x-h\right)}^2=4a\left(y-k\right)$ (1).

Substitute $\left(x,y\right)=\left(4,2\right)$ and $\left(h,k\right)=\left(3,0\right)$ in equation (1) and obtain the value of a.

$\begin{array}{c}{\left(4-3\right)}^2=4a\left(2-0\right)\\ 1=8a\\ a=\frac{1}{8}\end{array}$

Thus, the equation of the parabola with vertex (0, 3) is ${\left(x-3\right)}^2=4\left(\frac{1}{8}\right)\left(y-0\right)$.

That is, $f\left(x\right)=2{\left(x-3\right)}^2$.

Consider the graph of the function g.

Here, the graph represents that an open downward parabola which passing through the points (−2, 2), (0, 1) and (1, −2.5).

The general form of the quadratic equation is $y=a{x}^2+bx+c$ (2)

Substitute the points (−2, 2) and (1, −2.5) in equation (2) and obtain the equations as given below.

If (x, y) = (−2, 2), then the equation is

$2=4a-2b+1$

$2a-b=\frac{1}{2}$ (3)

If (x, y) = (0, 1), then the equation is

$-2.5=a+b+1$

$a+b=-3.5$ (4)

Add the equations (3) and (4) and find the value of a.

$\begin{array}{c}6a=-6\\ a=-1\end{array}$

Substitute $a=-1$ in equation (3) and get the value of $b=-2.5$.

Thus, the equation of the graph g is $y=\left(-1\right)x^2+\left(-2.5\right)x+c$.

Substitute $(0,1)$ in equation (2) and get the value of $c=1$.

That is, $g\left(x\right)=-x^2-2.5x+1$.