Question

Pls someone help, ill give ya brainliest :D

Answer 1

Answer:

Part A) The x-intercepts are the points (-2.5,0) and (0.5,0)

Part B) The vertex is a minimum and the coordinates are (-1,-9)

Part C) see the explanation

Step-by-step explanation:

we have

$f(x)=4x^{2}+8x-5$

Part A: What are the x-intercepts of the graph of f(x)?

we know that

The x-intercepts are the values of x when the value of the function f(x) is equal to zero

so

For f(x)=0

$4x^{2}+8x-5=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4x^{2}+8x-5=0$  

so

$a=4\\b=8\\c=-5$

substitute in the formula

$x=\frac{-8(+/-)\sqrt{8^{2}-4(4)(-5)}} {2(4)}$

$x=\frac{-8(+/-)\sqrt{144}} {8}$

$x=\frac{-8(+/-)12} {8}$

$x_1=\frac{-8(+)12} {8}=0.5$

$x_2=\frac{-8(-)12} {8}=-2.5$

therefore

The x-intercepts are the points (-2.5,0) and (0.5,0)

Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex?

we know that

The quadratic equation represent a vertical parabola open upward (because the leading coefficient is positive)

therefore

The vertex represent a minimum

$f(x)=4x^{2}+8x-5$

Convert to vertex form

Factor 4

$f(x)=4(x^{2}+2x)-5$

Complete the square

$f(x)=4(x^{2}+2x+1)-5-4$

$f(x)=4(x^{2}+2x+1)-9$

Rewrite as perfect squares

$f(x)=4(x+1)^{2}-9$ ----> equation in vertex form

The vertex is the point (-1,-9)

Part C: What are the steps you would use to graph f(x)?

To graph f(x)

I need

1) The direction of the graph (Is a vertical parabola open upward)

2) The coordinates of the vertex (-1,-9) ( in this case is a minimum)

3) The x-intercepts  (-2.5,0) and (0.5,0)

4) The y-intercept

The y-intercept is the value of f(x) when the value of x is equal to zero

so

For x=0

$f(0)=4(0)^{2}+8(0)-5\\f(0)=-5$

The y-intercept is the point (0,-5)

Plot the points and draw the parabola

see the attached figure to better understand the problem