Question

Find all values of x (if any) where the tangent line to the graph of the given equation is horizontal
f(x)=-4^2-3x

not so much looking for a direct answer just wondering how i would solve this question

Answer 1

You'd start by finding the derivative of the equation first

Answer 2

Yes, the solutions to the first derivative are the points where are max/min (horizontal tangent lines) of the parent function.

Answer 3

Assuming this is a calculus question

Answer 4

okay the derivative is -3 then?

Answer 5

Yes

Answer 6

thats my answer?

Answer 7

Is there supposed to be an x in -4^2?

Answer 8

oopes i meant -8*x-3 is the derivative

Answer 9

What is the given equation, exactly?

Answer 10

f(x)=-4x^2-3x my bad

Answer 11

but the derivative of that is -8x-3

Answer 12

Then yes the answer is the solutions to -8x-3

Answer 13

but how would i find any solutions if i wasnt given any more information?

Answer 14

How much less can they provide? They've given you the equation, unless they want you to perform a proof with just variables maybe

Answer 15

i even entered none and it said incorrect sooo im really not sure

Answer 16

You need to find the vertex of this parabola, because that is the point where the slope will be 0.

Answer 17

how would i find the vertex?

Answer 18

I'd put the parabola in the standard form to identify the vertex easier. This one seems relatively easy to complete the square at first glance, so I'd give it a shot. Do you know how to do that?

Answer 19

no i dont:(

Answer 20

You have to factor the coefficient of the leading term first \[-4(x^2 + \frac{3}{4}x)\] \[-4[(x^2 + \frac{3}{4}x + \frac{9}{64}) - \frac{9}{64}] = -4[(x + \frac{3}{8})^2 - \frac{9}{64}] \] distribute to expand and get the standard form \[-4(x + \frac{3}{8})^2 + \frac{9}{16}\]So the vertex is (-3/8, 9/16)

Answer 21

so after we have found the vertex what do we do next?