Inimum value of y=4x^2+4x-35?

1 answer:

8 0

Hello there!

There are two ways to find the minimum value of this function, but before I show you how, I am going to teach you a little bit about minimum value.

The minimum value on a parabola is the vertex or turning point. This means that the slope of the tangent line is horizontal, having a slope of 0.

The algebraic way to find the minimum or maximum value on a parabola is to use the formula -b/2a. Let's do it...

y=4x^2+4x-35
where a=4, b=4, and c=-35
-b/2a
-4/2(4)
-4/8
-1/2

Now let me show you the other way...
Take the derivative of
y=4x^2+4x-35
y'=8x+4
And set it equal to 0...
8x+4=0
8x=-4
x=-1/2

We got the same answer. Now that we know x=-1/2, plug this into the original equation to find y.
y=4x^2+4x-35
y=4(-1/2)^2+4(-1/2)-35
y=-1-2-35
y=-38

So the minimum point on this parabola is (-1/2,-38)

I really hope th

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