Find the vertex of the parabola.
y = -4x2 - 16x - 11
y = -4x^2 - 16x - 11
there are many ways to determine the vertex, but the easiest (for me) i think is to use derivatives. first we get the derivative with respect to x:
y = -4x^2 - 16x - 11
y' = -8x - 16
then we equate this to zero (because vertex is a maximum or a minimum and therefore the slope is zero):
0 = -8x - 16
8x = -16
x = -2
substitute this back to the original given equation:
y = -4x^2 - 16x - 11
y = -4(-2)^2 - 16(-2) - 11
y = -16 + 32 -11
y = 5
therefore vertex is at
(-2, 5)
hope this helps~ :)
If you don't know Calculus, like in Jai's method,
then....
the x value of the vertex is -b/(2a) = 16/-8 = -2
sub into original to get
y = -16 + 32 - 11 = 5
vertex is (-2,5)
To find the vertex of the parabola given by the equation y = -4x^2 - 16x - 11, we can use the formula -b/2a, where a and b are the coefficients of x^2 and x, respectively.
In this case, the coefficient of x^2 is -4, and the coefficient of x is -16.
Using the formula, we substitute these values into the formula:
x = -(-16) / (2 * (-4))
x = 16 / (-8)
x = -2
Now that we have the x-coordinate of the vertex, we can substitute this value back into the original equation to find the y-coordinate.
y = -4(-2)^2 - 16(-2) - 11
y = -4(4) + 32 - 11
y = -16 + 32 - 11
y = 5
Therefore, the vertex of the parabola is (-2, 5).