# What is the vertex for y = 8x^2 + 4x – 3?

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Given y= 8x^2 + 4x - 3

We need to find the vertex.

We know that:

a= 8 b= 4 c = -3

Let V be the vertex such that: V (xv, yv)

xv = -b/ 2a ==> -4/2*8 = -4/16 = -1/4

yv = -delta/ 4a = -(b^2 - 4ac)/4a = -(16+4*3*8)/4*8 = -112/32 = -7/2

Then the vertex is :

**V ( -1/4, -7/2) **

The vertex of a quadratic expression y = ax^2 + bx + c is given by ((-b/2a), (4ac – b^2)/4a)

We have y = 8x^2 + 4x – 3

Here a = 8, b = 4 and c = -3

–b/2a = -4 / 2*8 = -1/4

(4ac – b^2)/4a = (4*8*-3 – 4^2)/4*8

= (-96 – 16)/32

= -112/32

= -7/2

**Therefore the vertex is (- 1/4, -7/2).**

` y = 8x^2 + 4x – 3`

In order to find the vertex we need to know a,b and c

`a=8 ` (number in front of x^2) `b=4 ` (number in front of x) `c=-3` (number with no variable)

to find the x value you just need to use the formula `-b/(2a)`

`x= -4/(2(8))` `x=-4/16` simplify it `x=-1/4`

now plug in the x

`y = 8(-1/4)^2 + 4(-1/4) – 3`

`y=8(1/16)-1-3`

`y= (8/1) xx (1/16)= 8/16` simplified `= 1/2`

`y=1/2 - 4 ` you have to have the same denominator so set up -4 as a fraction:

`-4/1` and multiply by `2= -8/2`

`y=1/2 - 8/2 = -7/2`

vertex is `(-1/4 , -7 /2)`

The vertex of a parabola is the extreme point of the function. To determine the extreme point, we'll determine the critical points that are the roots of the first derivative of the function.

y = 8x^2 + 4x – 3

But y = f(x)

We'll calculate the first derivative of f(x), with respect to x.

f'(x) = 16x + 4

Now, we'll put f'(x) = 0, to determine the critical point:

16x + 4 = 0

We'll simplify by 4:

4x + 1 = 0

We'll subtract 1:

4x = -1

x = -1/4

The critical point of f(x) is x = -1/4. The extreme point of f(x), namely the vertex of the parabola, is f(-1/4).

f(-1/4) = 8/16 + 4*(-1/4) – 3

f(-1/4) = 1/2 - 1 - 3

f(-1/4) = 1/2 - 4

f(-1/4) = -7/2

**The coordinates of the vertex are: V(-1/4 , -7/2).**