Chapter 4.7, Problem 2E

a

To determine

To graph: the quadratic equation as mentioned in the problem.

Explanation of Solution

Given information:

The quadratic equation is $y=0.5x^2+2x-2$ .

Graph:

The graph mentioned below is in real number line of x axis and y axis.

Where vertex $\left(-10,-60\right)$ .

  

Interpretation:

Consider the quadratic equation $y=0.5x^2+2x-2$ .

The mentioned quadratic equation above will be in vertex form $y=a{\left(x-h\right)}^2+k$ where $\left(h,k\right)$ is the vertex.

We can write the mentioned quadratic equation in the vertex form as $\begin{array}{l}y=0.5x^2+2x-2\Rightarrow y=0.5\left(x^2+\frac{2}{0.5}x-\frac{2}{0.5}\right)\Rightarrow y=0.5\left(x^2+20x-20\right)\Rightarrow y=0.5\left(x^2+2*x*10+{10}^2-{10}^2-20\right)\\ \Rightarrow y=0.5\left({\left(x+10\right)}^2-{10}^2-20\right)\Rightarrow y=0.5\left({\left(x+10\right)}^2-120\right)\Rightarrow y=0.5{\left(x+10\right)}^2-60.\end{array}$

Therefore the vertex form of the equation is $y=0.5{\left(x+10\right)}^2-60$ .

Which is comparable to $y=a{\left(x-h\right)}^2+k$ where $a=0.5$ which is greater than $0$ .

Since $a>0$ the direction of graph will be positive side of y axis and the vertex is $\left(-10,-60\right)$ .

The discriminant of standard form of quadratic equation $y=0.5x^2+2x-2$ is $D=b^2-4*a*c\Rightarrow D={\left(2\right)}^2-4*\left(0.5\right)*\left(2\right)=4-4=0.$ .

Here $D=0$ .

Therefore the graph will lie on x axis.

At $x=0$ the $y=0.5x^2+2x-2$ will yield $y=-2$ .

Therefore graph will cut y axis at $\left(0,-2\right)$ .

Therefore the graph will be as mentioned below.

  

b

To determine

To draw: the table for the equation $y=0.5x^2+2x-2$ .

Answer to Problem 2E

The table of the equation is

  $\begin{array}{cc}x& y\\ -5& 0.5\\ -4& -2\\ -3& -3.5\\ -2& -4\\ -1& -3.5\\ 0& -2\\ 1& 0.5\end{array}$ .

Explanation of Solution

Given information:

The equation is $y=0.5x^2+2x-2$ and the value of $x$ is $-5,-4,-3,-2,-1,0,1.$

Formula:

We will substitute the value of $x$ in the equation $y=0.5x^2+2x-2$ .

Calculation:

Consider the equation $y=0.5x^2+2x-2$ .

Since the value of $x$ is $-5,-4,-3,-2,-1,0,1.$

Recall that we will substitute the value of $x$ in the equation $y=0.5x^2+2x-2$ .

Therefore

At $x=-5\Rightarrow y=0.5\left(-5*-5\right)+2\left(-5\right)-2=0.5$ .

At $x=-4\Rightarrow y=0.5\left(-4*-4\right)+2\left(-4\right)-2=-2$ .

At $x=-3\Rightarrow y=0.5\left(-3*-3\right)+2\left(-3\right)-2=-3.5$ .

At $x=-2\Rightarrow y=0.5\left(-2*-2\right)+2\left(-2\right)-2=-4$ .

At $x=-1\Rightarrow y=0.5\left(-1*-1\right)+2\left(-1\right)-2=-3.5$ .

At $x=0\Rightarrow y=0.5\left(0*0\right)+2\left(0\right)-2=-2$

At $x=1\Rightarrow y=0.5\left(-1*-1\right)+2\left(-1\right)-2=\mathrm{0.5.}$

Therefore the table will be as mentioned below.

  $\begin{array}{cc}x& y\\ -5& 0.5\\ -4& -2\\ -3& -3.5\\ -2& -4\\ -1& -3.5\\ 0& -2\\ 1& 0.5\end{array}$