Chapter 4.7, Problem 22PPS

To determine

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

Explanation of Solution

Given information:

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

Graph:

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

Where vertex $V\left(5,3\right)$ .

  

Interpretation:

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

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 $V\left(5,3\right)$ .

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

  $y={\left(x-5\right)}^2+3\Rightarrow y=x^2+25-10x+3\Rightarrow y=x^2-10x+3$ .

which is comparable to $a{x}^2+bx+c$ where $a=1$ 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 $y={\left(x-5\right)}^2+3$ is $D=b^2-4*a*c\Rightarrow d={\left(-10\right)}^2-4*\left(1\right)*\left(28\right)=100-112=-12$ .

Here $D<0$ .

Therefore the graph will lie below x axis.

At $x=0$ the $y={\left(x-5\right)}^2+3$ will yield $y=28$ .

Therefore graph will cut y axis at $\left(0,28\right)$ .

Therefore the graph will be as mentioned below.

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