calculusderivativeslimits
How would I solve the following?
Find an equation for the tangent line to the graph
$f(x)=3x^2-4x$ at the point $x=-1$
What I did is
$3(-1+h)^2-4(-1+h)-7)$
$3-6h+3h^2+4-4h-7)$
As h approaches zero $(2h+3h^2)/h$ limit equal $2$
then the equation I wrote is $y-7=2(x+1)$
But I am not sure if it is correct.
Simplify $f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$
And find where $f'(x) = 0$
$3x^2 - 6x - 9 = 0 \iff x^2 - 2x - 3 = (x + 1)(x - 3) = 0$
That's where slope is 0, hence any line tangent at that point will be horizontal: when $x = 3$ or when $x = -1$.
So the roots (x values) of the points you need are
$x_1 = 3$, and
$x_2 = -1$.
Then find the corresponding $y$ value in the ORIGINAL equation: $$\;f(x)=(x-2)(x^2-x-11),\;$$ to determine the two points: $(x_1, y_1), (x_2, y_2)\;$ where the line tangent to $f(x)$ is horizontal.
$y_1 = f(3) = 1 \cdot -5 = -5$
$y_2 = f(-1) = (-3)(-9) = 27$
So your points are $(3, -5)$ and $(-1, 27)$.
I've included a graph of the function $\;f(x)=(x-2)(x^2-x-11)\;$(in blue), along with the two horizontal lines tangent to the function: $\;y = -5,\;\; y = 27$ (violet and "brown", respectively) to see in a "picture" what is happening here.
The slope of a line perpendicular to $y = mx + b$ is equal to $-\frac 1m$, where slope $m$ of the equation $y = 3x - 1$ is $m = 3$.
So solve for the points at which the first derivative is equal to $-\dfrac 1m = -\dfrac 13$. You'll get a quadratic equation, from which you can find the two solutions.
$$y'=\frac{x-4}{(x-3)^2} = -\dfrac 13$$
$$ (x - 3)^2 = -3( x- 4),\;\;x \neq 3$$ $$x^2 - 6x + 9 = -3x + 12$$ $$ \iff x^2 - 3x - 3 = 0$$
Those solutions, (solving for $x$), will give you the $x$-coordinate of the points at which the tangent lines to the curve are perpendicular to the given line. We can simply use the quadratic equation to find the solutions, $x_1$ and $x_2$.
To find the corresponding $y$ coordinate, just plug each solution $x_i$ into the original curve (not into $y'$) and compute $y$.
Best Answer
You're approach is sound: the one (simple algebraic) mistake is that
$$(3-6h+3h^2+4-4h-7)\, =\, 3h^2 - 10h,\; \text{ and not} \;\;3h^2 + 2h:$$
So we need to evaluate the limit of $\;\;\dfrac{(3h^2 -10h)}{h}\; = \;\dfrac{h(3h -10)}{h}\;$ as $\;h\to 0:$
$$ \lim_{h\to 0}\,\frac{h(3h - 10)}{h} \; =\;\lim_{h\to 0}\, (3h-10) = -10.$$
So you need only replace "$2$" with "$-10$" in your equation of the line tangent to $\;f(x)=3x^2-4x\;$ at the point $\;(-1, 7)\;$, which gives you:
$$y - 7 = {-10}(x +1) \;\iff \;y = -10x-3$$