Equations of the tangent that pass through a point

**youngb11** · February 12th 2011, 10:49 AM

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

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 $(x,f(x))$ and $(2,3)$ which which would leave the slope to be $(2x^2)/(2-3)$

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

$x=0, 4/3$

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

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

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

**adkinsjr** · February 12th 2011, 10:59 AM

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

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

You should obtain a quadratic from this.

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

**youngb11** · February 12th 2011, 11:01 AM

Originally Posted by adkinsjr

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

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

You should obtain a quadratic from this.

Wouldn't $y_2$ be $2x^2+3$ ?

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?

**adkinsjr** · February 12th 2011, 11:06 AM

If that were the case, then the denominator would also change to $x-2$ . Here I'm defining the points $(x_1,y_1)=(x,2x^2+3)$ and $(x_2,y_2)=(2,3)$ . 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 $-1$ to get $\frac{2x^2}{x-2}$

**youngb11** · February 12th 2011, 11:10 AM

Originally Posted by adkinsjr

If that were the case, then the denominator would also change to $x-2$ . Here I'm defining the points $(x_1,y_1)=(x,2x^2+3)$ and $(x_2,y_2)=(2,3)$ . 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 $-1$ to get $\frac{2x^2}{x-2}$

Okay, so what would you do next in regards to the original question? Would you agree the first tangent's equation is $y=0$ ? If so, do you know how they got the second equation?

**adkinsjr** · February 12th 2011, 11:28 AM

I solved the quadratic:

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

$x(8-2x)=0$

$x=0$ and $x=4$

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

So just use the standard procedure. Find the derivative:

$f'(x)=4x$

The lines will have the form:

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

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

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

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

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

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

$y=3$

For the other equation I get:

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

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

**youngb11** · February 12th 2011, 11:37 AM

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

Thanks again!

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