Chapter 4.7, Problem 26PPS

To determine

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

Explanation of Solution

Given information:

The quadratic equation is $y=-3{\left(x-5\right)}^2-2$ .

Graph:

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

Where vertex $V\left(5,-2\right)$ .

  

Interpretation:

Consider the quadratic equation $y=-3{\left(x-5\right)}^2-2$ .

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 $V\left(5,-2\right)$ .

We can write the mentioned quadratic equation in the standard form as $y=-3x^2+30x-77$ which is comparable to $a{x}^2+bx+c$ where $a=-3$ which is less than $0$ .

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

The discriminant of standard form of quadratic equation $y=-3x^2+30x-77$ is $D=b^2-4*a*c\Rightarrow d={\left(30\right)}^2-4*\left(-3\right)*\left(-77\right)=900-924=-24$ .

Here $D<0$ .

Therefore the graph will lie below x axis.

At $x=0$ the $y=-3x^2+30x-77$ will yield $y=-77$ .

Therefore graph will cut y axis at $\left(0,-77\right)$ .

Therefore the graph will be as mentioned below.