Question

Determine the whether the quadratic equation below has a maximum or minimum. y = 4x² – 3x - 43

Answer 1

Answer:

Minimum

Need to know:

Derivative power rule: d/dx xⁿ = nxⁿ⁻¹

Derivative constant rule: d/dx c = 0

Step-by-step explanation:

A way to determine whether the quadratic equation has a maximum or a minimum is by graphing it.

If we look at the image below, we notice that the graph does not go to the bottom infinitely, but it keeps going up infinitely. This means there is a minimum to this graph.

Another way to determine this is through calculus.

First, we have to find the point of extremum

Find the derivative of the given equation

d/dx 4x² -  3x - 43

You can separate the terms and solve them individually

d/dx 4x² = 4(2)(x²⁻¹) = 8x

d/dx -3x = -3(1)(x¹⁻¹) = -3(1)(1) = -3

d/dx -43 = 0

y' = 8x - 3

Set y' to equal 0

8x - 3 = 0

Add 3 to both sides

8x - 3 = 0

    + 3   + 3

8x = 3

Divide both sides by 8

8x/8 = 3/8

x = 3/8

There is only one point of extremum

Now we have to test whether the equation changes from negative to positive or positive to negative

If it changes from negative to positive, that point is a minimum

If it changes from positive to negative, that point is a maximum

To find this, pick a number less than 3/8 and plug it in the place of x in the derivative equation. We'll use 0.

y' = 8(0) - 3 = -3

Now we will do the same for a number larger than 3/8. We'll use 1

y' = 8(1) - 3 = 5

Since the lesser side of 3/8 is negative and the larger side of 3/8 is positive, that means it changes from negative to positive. This point is a minimum.

Answer 2

A minimum because the function is positive. Positive quadratic functions always concave upwards (in a “bowl-like” shape), creating a curve/minimum at the bottom.