Find the vertex of y=5-2x-x^2?

Question

I got a different answer than someone in my class, a second opinion would be nice. Thanks!

Answer 1

If you are in a pre=calculus course you must complete the square for
y = -(x^2 + 2x -5) by adding and subtracting 1 to get

y = (x+1)^2 - 6 with vertex at (-1, -6)

If you are in calculus tae the derivative and set it to zero and solve to find the minimum value which is at the vertex

-2 -2x = 0 so x = -1 and f(-1) = - 6

which is what we found by the first method.

You can also just memorize the vertex is at x = -b/2a (from the quadratic formula, which gives

x = -(-2)/2(-1) = -1 and f(-1) = -6 as before.

Learn and understand how to get it one of the first two ways.

Answer 2

The function is a downward-shaped parabola. To find the vertex (maximum), take the 1st derivative of the function and set it equal to 0.....

y ' = -2 - 2x = 0

2x = -2 ------------------> x = -1

The vertex occurs when x = -1. Now, substitute this value back into the original function.....

y = 5 - 2(-1) - (-1)^2 = 5 + 2 - 1 = 6

Functions vertex is at point (-1, 6)