Answer :
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
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