Chapter 4.7, Problem 28PPS

To determine

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

Explanation of Solution

Given information:

The quadratic equation is $y=2x^2+10$ .

Graph:

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

Where vertex $V\left(0,10\right)$ .

  

Interpretation:

Consider the quadratic equation $y=2x^2+10$ .

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

Therefore the vertex of equation is $V\left(0,10\right)$ .

We can write the mentioned quadratic equation in the standard form as $y=2x^2+0x+10$ which is comparable to $a{x}^2+bx+c$ where $a=2$ which is greater than $0$ .

Since $a>0$ the direction of graph will be positive side of y axis.

The discriminant of standard form of quadratic equation $y=2x^2+0x+10$ is $D=b^2-4*a*c\Rightarrow d={\left(0\right)}^2-4*\left(2\right)*\left(10\right)=0-80=-80$ .

Here $D<0$ .

Therefore the graph will lie above x axis.

At $x=0$ the $y=2x^2+0x+10$ will yield $y=10$ .

Therefore graph will cut y axis at $\left(0,10\right)$ .

Therefore the graph will be as mentioned below.