A generic point on the graph of the curve has coordinates

The derivative gives us the slope of the tangent line at a given point:

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through
and have slope 
So, we can write its equation using the point-slope formula: a line with slope m passing through
has equation

In this case,
and
, so the equation becomes

We can rewrite the equation as follows:

We know that this function must give 0 when evaluated at x=0:

This equation has no real solution, so the problem looks impossible.
You have to understand the problem clearly before trying to answer the question. Only thing that needs to be done to answer the question is adding 5 with f(x). This will yield the required result.
g(x) = f(x) + 5
= (4x^2 - 8) + 5
= 4x^2 - 8 + 5
= 4x^2 - 3
I hope that this is the answer that you were looking for and the answer has actually come to your desired help.
Answer:
19
Step-by-step explanation:
7x-8=6x+11
-6x -6x
x-8=11
+8 +8
x=19
Answer:
The correct awnser is D.
Step-by-step explanation:
rise/run= 3/3
3/3=1