How to rewrite the function 2x^2-7x+5 by completing the square

see below

Step-by-step explanation:

A.  f(x) = 4x^2 - 7x - 15

We need the numbers to multiply to- 15

1*15  and 3*5  

One of the terms multiplies by 4 (or 2)

Try 4 first

4*-3 +5 = -7   That works

-3+5 = -15

f(x) = (4x+5)(x-3)

B.  We want the x intercepts so set equal to 0

0 = (4x+5)(x-3)

Using  the zero product  property

0 = 4x+5    x-3 =0

-5 = 4x           x=3

-5/4 =x            x=3

C  End behavior

Let x be negative infinity

The function is dominate by 4x^2

f(-∞) = 4 ( -∞) ^2 = 4 (∞) = ∞  

At negative infinity the functions goes to infinity

Let x be  infinity

The function is dominate by 4x^2

f(∞) = 4 ( ∞) ^2 = 4 (∞) = ∞  

At  infinity the functions goes to infinity

Part D:

We know the zeros at 3 and -5/4

We know the vertex is at  halfway between the zeros

(3 -5/4) /2 =  7/4 /2 = 7/8

Since the parabola opens up (The coefficient of the x^2 term is positive), we know we have a minimum.

f(7/8) = 4 (7/8)^2 -7(7/8) -15 = 4(49/64) -1/8 -15 =-289/16

We know the minimum the zeros and the end behavior, we can graph the functions