Derivative given separate constant for given function

Question

I do not understand what is meant here.

I must find the equation of the tangent line to a graph at $x = 0$ given the following function:

$y = 3e^x − 7x$

How do I compute this problem?

Answer 1

So you're starting with $y=3e^x-7x$. The derivative will give you, at every point, the slope of the tangent line to that point.

So you first need to find the derivative.

$$\frac{dy}{dx} = 3e^x-7$$

When you plug in an x-value to this, it's going to give you the slope of the tangent line. So, we're trying to find the slope at $x=0$, so we plug in $0$ for $x$.

Then you're going to want to use the linearization formula:

$$y = f'(k)(x-k)+f(k)$$

Where k is the x-value where you're trying to find the tangent line.

So when you plug in $x=0$, that's giving you $f'(k)$, which is $-4$.

You should be able to plug in from there.

Answer 2

HINT

The slope of the tangent line at a point is given by the value of the derivative at the point. So since, in your function $$\frac{\mathrm{d}y}{\mathrm{d}x}=3e^{x}-7$$ The value at $x=0$ (which is the slope) is $-4$.