How to find minimum and maximum values of this function

y = -2x squared - 16x + 18

the answer is 50 (according to the answer key) but im not sure how to get that answer?

thanks so much! xx

The maximum value (as it's -2x^2), is where the vertex is. We can solve for that by completing the square.

y = -2x^2 - 16x + 18

y = -2(x^2 + 8x) + 18

y = -2(x^2 + 8x + 16 - 16) + 18

y = -2(x + 4)^2 + 18 -2(-16)

y = -2(x + 4)^2 + 50

So the maximum value is at y = 50.

Alternatively, and as a means to check one's answer, you could differentiate this equation:

y' = -4x - 16

Then set this to 0 (as the derived equation here is the slope of -2x^2 - 16x + 18), as you want where the slope, m, equals 0.

-4x - 16 = 0

x = -4.

Plugging this into the original equation:

-2(-4)^2 - 16(-4) + 18 = 50.

Have a wonderful day!

The given equation is:
y(x) = -2x² - 16x + 18
In order to determine what the maximum/minimum point is, we need to differentiate dy/dx to give us,
y'(x) = -4x - 16
Now let y'(x) = 0 (since the gradient at the maximum/minimum point is zero), this gives,
-4x - 16 = 0 ⇒ x = -4
Substituting x = -4 back into the original equation, you solve for y(x),
y(-4) = -2(-4)² - 16(-4) + 18 = 50
Thus,
(-4, 50) is a maximum or a minimum.
In order to determine weather it is a maximum or minimum point, we look at the first derivative:
y"(x) = -4 < 0 ⇒ Maximum
Therefore 50 is the maximum of y.

Hope this helps

y = - 2x² - 16x + 18

I assume that you are not required to do this using calculus, in which case you should use the method called Completing the Square.

y = -2(x² + 8x) + 18

y = -2[x² + 8x + (8/2)² - (8/2)²] + 18

y = -2(x² + 8x + 16 - 16) + 18

y = -2[(x + 4)² - 16] + 18

y = -2(x + 4)² + 32 + 18

y = -2(x + 4)² + 50

This will have a maximum value of 50 when -2(x + 4)² is zero, which occurs when x = -4.

x is 2 (16x2=32+18=50)