Chapter 2, Problem 9CR

To determine

Writing the Quadratic function into standard form and finding the vertex,axis of symmetry, x-intercept and draw its graph.

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=-x^2-6x+10$

Method:

By using properties of standard quadratic equation.

By using quadratic formula.

Calculation:

As we know, the standard form of quadratic equation is $a{x}^2+bx+c$

Here, the standard form of quadratic equation of the given function is $f\left(x\right)=-x^2-6x+10$

Now,by comparing given fuction with standard form we get,

  $\begin{array}{l}a=-1\\ b=-6\\ c=10\end{array}$

As we know,vertex of quadratic equation is $\left(-\frac{b}{2a},-\frac{D}{4a}\right)$

Putting the values of a,b we get the vertex as $\left(-3,19\right)$ .

For a quadratic equation in standard form, $y=a{x}^2+bx+c$ ,the axis of symmetry is a vertical line $x=-\frac{b}{2a}$ .

Here, the axis of symmetry is $x=-\frac{b}{2a}=-\frac{-6}{2\times \left(-1\right)}=-3$ .

For x-intercepts put $f\left(x\right)=0$

  $\begin{array}{l}-x^2-6x+10=0\\ \text{further}\text{solving}\text{it}\text{by}\text{quadratic}\text{formula}\text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a},\text{we}\text{get}\\ \Rightarrow x=-7.359,1.359\end{array}$

Therefore intercepts point are $\left(-7.359,0\right)\text{and}\left(1.359,0\right)$ .

Graph:

  

From graph we can clearly see that the vertex of given function is $\left(-3,19\right)$ and interceptss are

  $\left(-7.359,0\right)\text{and}\left(1.359,0\right)$ .