Question

Hello, I need some assistance with this homework question please for precalculusHW Q22

Answer 1

Given the function:

`f(x)=x^2+10x+22`

Let's find the real zeros.

The Zeros of the function are also the x-intercept.

To find the real zeros, input 0 for f(x) and solve for x using the quadratic formula:

`x=(-b\pm√(b^2-4ac))/(2a)`

Use the standard form to find the values of a, b, and c.

We have:

`\begin{gathered} ax^2+bx+c=0 \\ \\ x^2+10x+22=0 \end{gathered}`

Therefore, we have:

a = 1

b = 10

c = 22

Plug in values into the quadratic formula and solve for x.

We have:

`\begin{gathered} x=(-10\pm√(10^2-4(1)(22)))/(2(1)) \\ \\ \\ x=(-10\pm√(100-88))/(2) \\ \\ x=(-10\pm√(12))/(2) \\ \\ x=(-10\pm√(12))/(2) \end{gathered}`

Simplify:

`\begin{gathered} x=(-10-3.4641)/(20),\text{ }(-10+3.4641)/(20) \\ \\ x=−6.7321,\text{ -3.2679} \end{gathered}`

Therefore, the zeros are:

x = -6.7321, -3.2679

The zeros are the x-intercepts.

ANSWER:

A. The zeros and the x-intercepts are the same. They are: -6.7321, -3.2679