f(x)=x^2+8x - 4

## What is the minimum y value on the graph of y= f(x)

# 3 Answers

For this function since the a coefficient is > 1, then the parabola that the equation forms open up and its vertex is at a minimum point.

If this function is set to zero then the x intercepts for the graph can be found:

We proceed as follows: for y = 0 , Then x^{2} +8x- 4 = 0

Solving this quadratic form, x = (-b ± √ b^{2} - 4ac)/2a

a =1 , b = 8, c = - 4, by substitition x = ( -8 ± √ (8)^{2}-4(1)(-4)) / (2×1) and x = 1/2 , x = -8.5

the points (1/2 , 0) and ( -8.5,0) are the x intercepts of the graph . Knowing that the parabola opens up means that the minimum point is below the x axis . the y ordinate of this point is defenitely in the negative region of the Y axis. Since the vertex has only one x coordinate, then the only condition when the equation above has one root is when √ b2 - 4ac = 0 and x = -b/2a

x = -8/2(1) = -4 is the x coordinate of the vertex that lies in the third quadrant since the y ordinate is also negative.

To find y, substitute the value of x into the equation.

y = (-4)^{2}+ 8(-4) - 4 = -20

the minimum point is ( -4, -20) , this is the lowest point on parabola and its vertex

to test this minimum substitute x = -3 right of the vertex point , and x = -5 left of the vertex into the equation:

Y = (-3)^{2}+ 8(-3) - 4 = -19

Y = ( -5)^{2}+ 8(-5)- 4 = -19

The points on the graph are (-3, -19) , (-5 , -19) are both higher then the vertex point (-4,-20) and the point of the vertex is the minimum point the parabola openeing upward can attend.

1) Find the derivative of the function: f"(x) = 2x+8. Make it equal zero: 2x +8 = 0, thus x = -4.

This is a point of minimum because the parabola is opened up. Calculate the minimum of the function by plugging x = -4 into the function: f(-4) =(- 4)^{2} + 8x (-4) -4 = 16-32 - 4 =- 20.

2) Without derivatives. The minimum (or maximum) of the function is a vertex of the parabola. In very generic form if you have an equation for a parabola in the form ax^{2} + bx + c the location of the vertex on x - axis is defiend by the formula:
x = -b/2a. Because a = 1 and b = 8 we have x = -4. Then,

f(4) = -20 (minimum).

*Graph of the function y = ƒ(x) = ax ^{2} + bx + c (a≠0) is parabola.
*

*If a > 0 , then parabola is open up and function has minimum value.*

*If a < 0 , then parabola is open down and function has maximum value.*

y = ƒ(x) = x

^{2}+ 8x - 4

a = 1 > 0 , then y-coordinate of a vertex is a minimum of a given function.

Let's find x-coordinate of vertex.

x = - b / (2a)

x = - 8 / 2 = - 4

Now, let's "plug in" (-4) into original function, and we will find the y-coordinate of the vertex.

*= (-4)*

**y**^{2}+ 8 · (-4) - 4

**= - 20**