Question

Use the function f(x) = 5x^2 + 2x − 3 to answer the questions. Part A: Completely factor f(x). (2 points) Part B: What are the x-intercepts of the graph of f(x)? Show your work. (2 points) Part C: Describe the end behavior of the graph of f(x). Explain. (2 points) Part D: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part B and Part C to draw the graph. (4 points)

Answer 1

Answer:

see below

Step-by-step explanation:

f(x) = 5x^2 + 2x − 3

Factor

f(x) = (x + 1) (5 x - 3)

To find the x intercepts, set f(x) =0

0 = (x+1) ( 5x-3)

Using the zero product property

x+1 = 0  5x-3 =0

x=-1    5x=3

          x = 3/5

The x intercepts are (-1,0) and (3/5,0)

When x goes to -∞ the 5x^2 term dominates

f( -∞) = 5 (  -∞) ^2 = 5 ( ∞) =  ∞

As x  goes to  -∞     f(x) goes to  ∞

When x goes to ∞ the 5x^2 term dominates

f( ∞) = 5 (  ∞) ^2 = 5 ( ∞) =  ∞

As x  goes to  ∞     f(x) goes to  ∞

We need the end behavior, the x intercepts and the vertex to graph

We know this opens upward since the coefficient of x^2 is positive

The vertex is 1/2 way between the zeros

(-1+3/5) /2 = (-2/5) /2 = -1/5

Substitute this into f(x) to find the minimum

f(-1/5) = (-1/5 + 1) (5 *-1/5 - 3) = 4/5 * (-4) = -16/5

Answer 2

Answer:

Part A

$f(x)=5x^{2} +2x-3$

we multiply by 5

$(5x)^2+2(5x)-15$   and we search for two numbers that added give you 2 and multiplied 15, and those are 5 and -3, so that gives us

$(5x-3)(5x+5)$ but because we multiplied by 5, we have to divide by 5

$\frac{(5x-3)(5x+5)}{5}=\frac{5*(5x-3)(x+1)}{5}=(5x+3)(x+1)$

Part B.

Well we know that f(x) intercepts x when f(x)=0

so we have

$0=(5x-3)(x+1)$ so we have two cases

when $5x+3=0$ and when $x+1=0$

so we solve for them and get

$x_1=\frac{3}{5}$

$x_2=-1$

Part C.

well as I understand the end behavior is that in this function is that as x grows towards infinity f(x) also grows towars infinity because the function has a positive leading coefficient.

Part D.

Well we alredy know two points when f(x)=0.

we can also work out when x=0

$f(x)=(5x-3)(x+1)\\f(0)=(5(0)-3)(0+1)\\f(0)=-3$

To find the other points we need to find the vertex.

$f(x)=5x^2+2x-3\\f(x)=(5x^2+2x)-3\\f(x)=5(x^2+\frac{2x}{5} )-3\\f(x)=5(x+\frac{1}{5} )^2-3$

so the vertex is in x=-0.2

$f(0.2)=5(-0.2)^2+2(-0.2)-3\\f(0.2)=0.2-0.4-3\\f(0.2)=-3.2$

the vertex is in (-0.2,-3.2), and we can graph additional points by giving value to x