Chapter 2.4, Problem 90E

(a).

To determine

To find:

The $x$ intercepts of the function $f\left(x\right)$ and $g\left(x\right)$ .

Answer to Problem 90E

The $x$ intercepts in point form for $f\left(x\right)$ is $3\pm \sqrt{2}$ and the function $g\left(x\right)$ has no $x$ intercepts.

Explanation of Solution

Given:

  $f\left(x\right)=2{\left(x-3\right)}^2-4$ and $g\left(x\right)=-2{\left(x-3\right)}^2-4$

Concept Used:

To find the $x$ intercepts substitute in $0$ for $f\left(x\right)$ and $g\left(x\right)$ .

Calculation:

First solve for $f\left(x\right)$ .

  $\begin{array}{c}2{\left(x-3\right)}^2-4=0\\ 2\left(x^2-6x+9\right)-4=0\\ x^2-6x+9-2=0\\ x^2-6x+7=0\end{array}$

Solve the above by using quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1$ , $b=-6$ and $c=7$ in the above formula:

  $\begin{array}{c}x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(7\right)}}{2\left(1\right)}\\ =\frac{6\pm \sqrt{36-28}}{2}\\ =\frac{6\pm \sqrt{8}}{2}\\ =\frac{6\pm 2\sqrt{2}}{2}\\ =\frac{2\left(3\pm \sqrt{2}\right)}{2}\\ =3\pm \sqrt{2}\end{array}$

So, the $x$ intercepts in point form for $f\left(x\right)$ is $3\pm \sqrt{2}$ .

Now, solve for $g\left(x\right)$ .

  $\begin{array}{c}-2{\left(x-3\right)}^2-4=0\\ 2{\left(x-3\right)}^2+4=0\\ 2\left(x^2-6x+9\right)+4=0\\ x^2-6x+9+2=0\\ x^2-6x+11=0\end{array}$

Solve the above by using quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1$ , $b=-6$ and $c=11$ in the above formula:

  $\begin{array}{c}x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(11\right)}}{2\left(1\right)}\\ =\frac{6\pm \sqrt{36-44}}{2}\\ =\frac{6\pm \sqrt{-8}}{2}\\ =\frac{6\pm \sqrt{i^{2}8}}{2}\\ =\frac{6\pm 2i\sqrt{2}}{2}\\ =\frac{2\left(3\pm i\sqrt{2}\right)}{2}\\ =3\pm i\sqrt{2}\end{array}$

The solution is the $x$ value of the point of intersection but from the above solution roots are imaginary.

So, the function $g\left(x\right)$ has no $x$ intercepts.

Conclusion:

Hence, the $x$ intercepts in point form for $f\left(x\right)$ is $3\pm \sqrt{2}$ and the function $g\left(x\right)$ has no $x$ intercepts.

(b).

To determine

To find:

The values of $x$ for both the function $f\left(x\right)$ and $g\left(x\right)$ .

Answer to Problem 90E

The values of $x$ for $f\left(x\right)$ is $3\pm \sqrt{2}$ and the values of $x$ for $g\left(x\right)$ is $3\pm i\sqrt{2}$ .

Explanation of Solution

Given:

  $f\left(x\right)=2{\left(x-3\right)}^2-4$ and $g\left(x\right)=-2{\left(x-3\right)}^2-4$

Concept Used:

Apply quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

First solve for $f\left(x\right)$ .

  $\begin{array}{c}2{\left(x-3\right)}^2-4=0\\ 2\left(x^2-6x+9\right)-4=0\\ x^2-6x+9-2=0\\ x^2-6x+7=0\end{array}$

Solve the above by using quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1$ , $b=-6$ and $c=7$ in the above formula:

  $\begin{array}{c}x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(7\right)}}{2\left(1\right)}\\ =\frac{6\pm \sqrt{36-28}}{2}\\ =\frac{6\pm \sqrt{8}}{2}\\ =\frac{6\pm 2\sqrt{2}}{2}\\ =\frac{2\left(3\pm \sqrt{2}\right)}{2}\\ =3\pm \sqrt{2}\end{array}$

So, the values of $x$ for $f\left(x\right)$ is $3\pm \sqrt{2}$ .

Now, solve for $g\left(x\right)$ .

  $\begin{array}{c}-2{\left(x-3\right)}^2-4=0\\ 2{\left(x-3\right)}^2+4=0\\ 2\left(x^2-6x+9\right)+4=0\\ x^2-6x+9+2=0\\ x^2-6x+11=0\end{array}$

Solve the above by using quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1$ , $b=-6$ and $c=11$ in the above formula:

  $\begin{array}{c}x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(11\right)}}{2\left(1\right)}\\ =\frac{6\pm \sqrt{36-44}}{2}\\ =\frac{6\pm \sqrt{-8}}{2}\\ =\frac{6\pm \sqrt{i^{2}8}}{2}\\ =\frac{6\pm 2i\sqrt{2}}{2}\\ =\frac{2\left(3\pm i\sqrt{2}\right)}{2}\\ =3\pm i\sqrt{2}\end{array}$

So, the values of $x$ for $g\left(x\right)$ is $3\pm i\sqrt{2}$ .

Conclusion:

Hence, the values of $x$ for $f\left(x\right)$ is $3\pm \sqrt{2}$ and the values of $x$ for $g\left(x\right)$ is $3\pm i\sqrt{2}$ .

(c).

To determine

To explain:

The zeros of $f\left(x\right)$ and $g\left(x\right)$ are related to whether their graphs have $x$ intercepts.

Answer to Problem 90E

For the function $f\left(x\right)=2{\left(x-3\right)}^2-4$ whose solutions are real, but irrational then $x$ intercepts and zeros are same.

For the function $g\left(x\right)=-2{\left(x-3\right)}^2-4$ , whose solutions are imaginary, that means the function has no $x$ intercepts.

Explanation of Solution

Given:

  $f\left(x\right)=2{\left(x-3\right)}^2-4$ and $g\left(x\right)=-2{\left(x-3\right)}^2-4$

The $x$ intercepts are the points at which the parabola crosses $x$ axis. If they exist, the $x$ intercepts represent the zeros, or roots, of the quadratic function, the values of $x$ at which $f\left(x\right)=0$ .

For the function $f\left(x\right)=2{\left(x-3\right)}^2-4$ whose solutions are real, but irrational then $x$ intercepts and zeros are same.

For the function $g\left(x\right)=-2{\left(x-3\right)}^2-4$ , whose solutions are imaginary, that means the function has no $x$ intercepts.

Conclusion:

For the function $f\left(x\right)=2{\left(x-3\right)}^2-4$ whose solutions are real, but irrational then $x$ intercepts and zeros are same.

For the function $g\left(x\right)=-2{\left(x-3\right)}^2-4$ , whose solutions are imaginary, that means the function has no $x$ intercepts.

(d).

To determine

To explain:

Whether the graph of $f\left(x\right)=a{\left(x-h\right)}^2+k$ has $x$ intercepts, and whether the zeros of f are real or complex.

Answer to Problem 90E

If the value of $k$ is replaced with $-k$ then the function $f\left(x\right)$ has real roots and the function have $x$ intercepts.

Explanation of Solution

Given:

  $f\left(x\right)=a{\left(x-h\right)}^2+k$

The $x$ intercepts are the points at which the parabola crosses $x$ axis. If they exist, the $x$ intercepts represent the zeros, or roots, of the quadratic function, the values of $x$ at which $f\left(x\right)=0$ .

If the solutions are imaginary, that means the function has no $x$ intercepts.

  $f\left(x\right)=a{\left(x-h\right)}^2+k$ is a parabola in vertex form, the roots of the quadratic are given by the formula below. It gives the location on the $x$ axis of the two roots.

  $x=\pm \sqrt{\frac{-k}{a}}+h$

But the expression inside the square root is negative, there are no real roots.

So, solutions are imaginary, that means the function has no $x$ intercepts.

But, if the value of $k$ is replaced with $-k$ then the function $f\left(x\right)$ has real roots and the function have $x$ intercepts.

Conclusion:

If the value of $k$ is replaced with $-k$ then the function $f\left(x\right)$ has real roots and the function have $x$ intercepts.