Chapter 4.1, Problem 64HP

a.

To determine

Find the function with maximum value of $8$ .

Answer to Problem 64HP

  $f\left(x\right)=-x^2+8$

Explanation of Solution

Given information:

The table represents a quadratic function.

  $\begin{array}{cccccccccc}x& -20& c& -5& -1& d-1& 5& 7& 15& 14-c\\ y& -377& -13& -2& 22& a& a-24& -b& -202& -377\end{array}$

Calculation:

Assume $f\left(x\right)$ is a quadratic function with $x-coordinate$ of vertex zero and $y-coordinate$ of vertex $8$ unit, Then quadratic function of $f\left(x\right)$ is,

  $f\left(x\right)=-x^2+8$

The coefficient of $x^2$ is negative so the function is in parabolic form therefore the maximum value is attained at the $x-coordinate$ of the vertex of the parabola.

  $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{0}{2\left(-1\right)}\\ x=0\\ now\max imumvalueaty-coordinate,\\ putx=0\\ f\left(0\right)=-0^2+8\\ =8\end{array}$

Hence the quadratic function with maximum value $8$ is, $f\left(x\right)=-x^2+8$

b.

To determine

Find the function with minimum value of $-4$ .

Answer to Problem 64HP

  $f\left(x\right)=x^2-4$

Explanation of Solution

Given information:

The table represents a quadratic function.

  $\begin{array}{cccccccccc}x& -20& c& -5& -1& d-1& 5& 7& 15& 14-c\\ y& -377& -13& -2& 22& a& a-24& -b& -202& -377\end{array}$

Calculation:

Assume $f\left(x\right)$ is a quadratic function with $x-coordinate$ of vertex zero and $y-coordinate$ of vertex $-4$ unit,

Then quadratic function of $f\left(x\right)$ is,

  $f\left(x\right)=x^2-4$

The coefficient of $x^2$ is positive so the function is in parabolic form therefore the mainimum value is attained at the $x-coordinate$ of the vertex of the parabola.

  $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{0}{2\left(-1\right)}\\ x=0\\ now\max imumvalueaty-coordinate,\\ putx=0\\ f\left(0\right)=0^2-4\\ =-4\end{array}$

Hence the quadratic function with maximum value $-4$ is, $f\left(x\right)=x^2-4$

c.

To determine

Find the function with vertex $\left(-2,6\right)$ .

Answer to Problem 64HP

  $f\left(x\right)=x^2+4x+10$

Explanation of Solution

Given information:

The table represents a quadratic function.

  $\begin{array}{cccccccccc}x& -20& c& -5& -1& d-1& 5& 7& 15& 14-c\\ y& -377& -13& -2& 22& a& a-24& -b& -202& -377\end{array}$

Calculation:

Assume $f\left(x\right)$ is a quadratic function with $x-coordinate$ of vertex $-2$ and $y-coordinate$ of vertex $10$ unit,

Then quadratic function of $f\left(x\right)$ is,

  $f\left(x\right)=x^2+4x+10$

Now the vertex is $\left(-2,6\right)$

  $\begin{array}{l}f\left(-2\right)={\left(-2\right)}^2+4\left(-2\right)+10\\ =4-8+10\\ =6\end{array}$

Hence the quadratic function with vertex $\left(-2,6\right)$ is $f\left(x\right)=x^2+4x+10$ .