0

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

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

### 3 Answers by Expert Tutors

Roger N. | Client Satisfaction is PriorityClient Satisfaction is Priority
4.8 4.8 (50 lesson ratings) (50)
0

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 x2 +8x- 4 = 0

Solving this quadratic form,  x = (-b ± √ b2 - 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.

Grigori S. | Certified Physics and Math Teacher G.S.Certified Physics and Math Teacher G.S.
0

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 ax2 + 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).

Nataliya D. | Patient and effective tutor for your most difficult subject.Patient and effective tutor for your mos...
0

Graph of the function y = ƒ(x) = ax2 + 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) = x2 + 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.
y = (-4)2 + 8 · (-4) - 4 = - 20