Chapter 4.7, Problem 24PPS

To determine

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

Explanation of Solution

Given information:

The quadratic equation is $y=-2{\left(x-5\right)}^2+0$ .

Graph:

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

Where vertex $V\left(5,0\right)$ .

  

Interpretation:

Consider the quadratic equation $y=-2{\left(x-5\right)}^2+0$ .

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 $V\left(5,0\right)$ .

We can write the mentioned quadratic equation in the standard form as mentioned below

  $y=-2{\left(x-5\right)}^2+0\Rightarrow y=\left(-2x^2+20x-50\right)$ .

which is comparable to $a{x}^2+bx+c$ where $a=-2$ which is less than $0$ .

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

The discriminant of standard form of quadratic equation $y=-2x^2+20x-50$ is $D=b^2-4*a*c\Rightarrow d={\left(20\right)}^2-4*\left(-2\right)*\left(-50\right)=400-400=0$ .

Here $D=0$ .

Therefore the graph will lie on x axis.

At $x=0$ the $y=-2x^2+20x-50$ will yield $y=-50$ .

Therefore graph will cut y axis at $\left(0,-50\right)$ .

Therefore the graph will be as mentioned below.

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