# Equation of the tangent line to the curve

1. Aug 30, 2007

### kdpointer

1. The problem statement, all variables and given/known data

Find an equation of the tangent line to the curve: 7*x*e^(x)+8 at (0,8)

2. Relevant equations

Derivative I guess?

3. The attempt at a solution

I know you have to take the derivative of the equation given which I think is

7*x*e^(x) + 7*e^(x)

Then you plug it into slope intercept form: y-y=m(x-x)

I did this and got y-8=(7*e^(x)*(x+1))*x, but apparently that's not right...

What did I do wrong? Maybe i just typed it in wrong?

2. Aug 30, 2007

### learningphysics

Calculate the slope first by plugging in x=0 into the derivative equation... then when you have the number for the slope, plug it into the slope intercept form.

The slope should just be a number.

3. Aug 30, 2007

### Dick

Yeah, it's pretty wrong. In the line equation m is a constant. Otherwise, it wouldn't be a line, now would it? Put (0,8) into the derivative before you plug it into the line equation.

4. Aug 30, 2007

### rocomath

so essentially you have to use your x-intercept twice

5. Aug 30, 2007

### kdpointer

So the slope is just 7?

6. Aug 30, 2007

### kdpointer

Thus the final equation is y-8 = 7x ?

7. Aug 30, 2007

### rocomath

it should probably be in slope-intercept form

y=mx+b

8. Aug 30, 2007

### kdpointer

It says that point-slope form is fine.. but it would be y=7x+8

9. Aug 30, 2007

### learningphysics

Yup. That's right.

10. Aug 30, 2007

### kdpointer

Alright.. thanks a lot!