Question

find the tangent line to the curve

y = -3x^2+5x-1

(0,1)

Answer 1

Here's what I did

Answer 2

Took the derivative of the function and then evaluated it at the point x = 0

\[\frac{ dy }{ dx }(-3x^{2}+5x-1) = -6x+5\]

\[|_{x=0} -6(0)+5 = 5 \]

they said equation I assume i'm done here right?

Answer 3

You found the slope of the tangent line.

Tangent line is an equation of a line.

Answer 4

You have the slope, and the point the tangent line passes through.

Answer 5

that's not the same as the y intercept right?

Answer 6

Use y=mx+b or y-y1 = m(x-x1).

You have the slope, m. And the point the line passes through. The point happens to be the y intercept in this case.

Answer 7

If \(y\) is the tangent line to the function \(f\) at \(x=a\), then: \[
\frac{y(x)-y(a)}{x-a} =f'(a)
\]

Answer 8

okay, so in this case it would be y =5x+1right?

is that the case in every situation?

Answer 9

Your slope is \(\color{#000000}{ \displaystyle f'(0) }\), correct?
And your points is \(\color{#000000}{ \displaystyle (0,f(0)) }\)

(well, you know f(0)=1, but I will write it ge nerally)

So, the equation of the tangent line is:
\(\color{#000000}{ \displaystyle y-y_1=m(x-x_1)\quad \Longrightarrow \quad y-f(0)=f'(0)\cdot (x-0) }\)

Answer 10

That is just an easy use of a point slope form.

Answer 11

yep

Answer 12

They don't always give you the y-intercept. But when they do, they prefer Dos Equis.

Answer 13

Yes, you can certainly write the equation like you want, but if you don't want to preform any algebra in your head and had a hard day, then there you go ... ))

Answer 14

perform**

Answer 15

interesting thanks!

Answer 16

lol, yw.

Answer 17

Eh, I my opinion the easiest thing you can do is solve for \(y\) given: \[
\frac{y-f(x_0)}{x-x_0} = f'(x_0)
\]To me it's the easiest, most straight forward approach.

Answer 18

OS keeps acting up for me will look this over!