# Thread: Equations of the tangent that pass through a point

1. ## Equations of the tangent that pass through a point

I'm trying to determine the tangents to the curve $\displaystyle f(x)=2x^2+3$ that pass through the point $\displaystyle (2,3)$

I got the first equation right, but the second one is apparently wrong (according to the book).

What I did was set the derivative equal to the slope of two points. Points I used were $\displaystyle (x,f(x))$ and $\displaystyle (2,3)$ which which would leave the slope to be $\displaystyle (2x^2)/(2-3)$

$\displaystyle 4x=(2x^2)/(2-3)$
$\displaystyle 0=2x(3x-4)$

$\displaystyle x=0, 4/3$

First equation is the tangent that passes through $\displaystyle f(0)=3$ so $\displaystyle (0,3)$. Finding the equation by setting the derivative at $\displaystyle f(0)$ equal to the slope like this:
$\displaystyle 0=(y-3)/(x-0)$
$\displaystyle y=3$

But I'm getting a different answer for the second equation:/

The book has the second equation as: $\displaystyle 16x-y-29=o$

2. I think there's an error in your slope calculation. The slope of the tangent line through the points $\displaystyle (2,3)$ and $\displaystyle (x,2x^2+3)$ is $\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{3-(2x^2+3)}{2-x}$. The derivative of the function is $\displaystyle 4x$, and should be equal to slope at the corresponding value of x. So you have:

$\displaystyle 4x=\frac{-2x^2}{2-x}$

You should obtain a quadratic from this.

Edit: Note there should be two tangents, since you'll probably find two real roots.

I think there's an error in your slope calculation. The slope of the tangent line through the points $\displaystyle (2,3)$ and $\displaystyle (x,2x^2+3)$ is $\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{3-(2x^2+3)}{2-x}$. The derivative of the function is $\displaystyle 4x$, and should be equal to slope. So you have:

$\displaystyle 4x=\frac{-2x^2}{2-x}$

You should obtain a quadratic from this.
Wouldn't $\displaystyle y_2$ be $\displaystyle 2x^2+3$?

Regardless, that's pretty much what I did and ended up with the roots of the quadratic as 0 and 4/3. Can you tell me what I would do next?

4. If that were the case, then the denominator would also change to $\displaystyle x-2$. Here I'm defining the points $\displaystyle (x_1,y_1)=(x,2x^2+3)$ and $\displaystyle (x_2,y_2)=(2,3)$. It's an arbitrary decision since we're only interested in the change between the values. You get the same slope either way. Just muliply by $\displaystyle -1$ to get $\displaystyle \frac{2x^2}{x-2}$

If that were the case, then the denominator would also change to $\displaystyle x-2$. Here I'm defining the points $\displaystyle (x_1,y_1)=(x,2x^2+3)$ and $\displaystyle (x_2,y_2)=(2,3)$. It's an arbitrary decision since we're only interested in the change between the values. You get the same slope either way. Just muliply by $\displaystyle -1$ to get $\displaystyle \frac{2x^2}{x-2}$
Okay, so what would you do next in regards to the original question? Would you agree the first tangent's equation is $\displaystyle y=0$? If so, do you know how they got the second equation?

$\displaystyle 8x-4x^2=-2x^2$

$\displaystyle x(8-2x)=0$

$\displaystyle x=0$ and $\displaystyle x=4$

Basically we've reduced the problem down to: find the equation of the tangent line to the curve $\displaystyle f(x)=2x^2+3$ at $\displaystyle x=0$ and $\displaystyle x=4$.

So just use the standard procedure. Find the derivative:

$\displaystyle f'(x)=4x$

The lines will have the form:

$\displaystyle y-y_o=f'(x_o)(x-x_o)$

Evaluate the derivative at $\displaystyle x=0$ and $\displaystyle x=4$

$\displaystyle f'(0)=0$ and $\displaystyle f'(4)=16$

The points, which will be our $\displaystyle (x_o,y_o)$, on the curve are $\displaystyle (0, f(0))=(0,3)$ and $\displaystyle (4,f(4))=(4,35)$

The first one is easy because the slope is zero, so it's a constant:

$\displaystyle y-3=0(x-0)$

$\displaystyle y=3$

For the other equation I get:

$\displaystyle y-35=16(x-4)$

$\displaystyle y=16x-64+35$
$\displaystyle y=16x-29$

7. Ah, thanks a lot for the very detailed explanation! I see where I went wrong.

Thanks again!