148k views
4 votes
Hi, can you help me to solve exercise #2 please

Hi, can you help me to solve exercise #2 please-example-1
User Lohith
by
8.4k points

1 Answer

6 votes

N 2

we have

f(x)=-x^2-4x+12

this is a vertical parabola open downward (because the leading coefficient is negative)

The vertex represents the maximum

Convert the given equation into vertex form

f(x)=a(x-h)^2+k

where

(h,k) is the vertex

The axis of symmetry is equal to x=h

so

f(x)=-x^2-4x+12

factor -1

f(x)=-(x^2+4x)+12

Complete the square

f(x)=-(x^2+4x+2^2-2^2)+12

f(x)=-(x+2)^2+16

the vertex is (-2,16)

The axis of symmetry is x=-2

Find out the y-intercept (value of f(x) when the value of x=0)

For x=0

f(x)=-(0)^2-4(0)+12

f(x)=12

The y-intercept is (0,12)

Graph the function

we have the points

(-2,16) vertex

(0,12) y-intercept

Find out te x-intercepts (values of x when the value of f(x)=0

-x^2-4x+12=0

solve the quadratic equation using the formula

a=-1

b=-4

c=12

substitute


x=\frac{-(-4)\pm\sqrt[]{-4^2-4(-1)(12)}}{2(-1)}
x=(4\pm8)/(-2)

The values of x are

x=

User Nathan Feger
by
7.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories