Question

Let f(x) = 3 x^2 - 8.
Find an equation of the line tangent to the graph of f at the point where x = -3.

y =

Answer 1

Do you know that the first derivative is the tangent function?

Answer 2

the power rule?

Answer 3

Yes

Answer 4

f'(X)= rx^r-1

Answer 5

f'(x)=6x

Answer 6

that the deriv 6x so plug that in the equation

Answer 7

At x = -3, f'(x)=-18 and that is the slope of the tangent at x = -3

Answer 8

y=mx+b right? or y-y1=mx-x1

Answer 9

y=6x That is the equation of the tangent function. At x = -3, it has a value of -18 and that is the slope of the tangent at that point.

Answer 10

Why don't you draw the graph of the parabola and then draw the tangent at x = -3 and see what it looks like?

Answer 11

That would help you understand.

Answer 12

wait what is the full equation y=6x?

Answer 13

So if you want the equation of the tangent you know its slope is -18 and you know it goes through the point (-3,19) So use either y= mx+b or point slope form. Whichever you prefer.

Answer 14

y=6x is the function that gives you the slope of the tangent at any x value of the parabola.

Answer 15

y=-18(-3)+b

Answer 16

\[y-19=-18(x+3)\]
\[y=-18x-35\]

Answer 17

y=-18(-3)+b
19=-18(-3)+b

Answer 18

for some reason im not writing it corrrectly bcause my book says its wrong

Answer 19

what does your book say it is?

Answer 20

equation of the tangent line of y = x3 at x = 2

Answer 21

3 x^2 = f(x)+8

Answer 22

We are not working with the function y = x^3. We are doing y = 3x^2-8

Answer 23

What does your book say for this one?

Answer 24

it says y=-18x+b

Answer 25

For the final answer?

Answer 26

yeah they only want the equation

Answer 27

of a line

Answer 28

Well it would be easy to find b and if i were you I would give both answers.

Answer 29

ok thanks

Answer 30

Because the problem specifies to find the equation of the tangent at x = -3.

Answer 31

yw