# Thread: Help with Tangent lines and Slope Predictor

1. ## Help with Tangent lines and Slope Predictor

There are two lines that pass through the point (5,10) and are tangent to the parabola

f(x) = -16+10x - x^2 .

Find the equations for both of them.

*So far I know that i should have something like

m(a) = [ f(a) -10 ] / [ a - 5 ] = f'(a)

and f'(a)= -2x+10
right? so how do I solve for a? I think that's what i'm supposed to be doing. and yeah I know also the derivative is f'(a)= -2x+10

2. Originally Posted by pseizure2000
There are two lines that pass through the point (5,10) and are tangent to the parabola

f(x) = -16+10x - x^2 .

Find the equations for both of them.

*So far I know that i should have something like

m(a) = [ f(a) -10 ] / [ a - 5 ] = f'(a)

and f'(a)= -2x+10
right? so how do I solve for a? I think that's what i'm supposed to be doing. and yeah I know also the derivative is f'(a)= -2x+10
we know that the lines must be of the form y = mx + b, right? since they pass through the point (5,10), we have by the point-slope form that te lines are of the form:

$y - 10 = m(x - 5)$

$\Rightarrow y = mx + (10 - 5m)$

Now, since the lines are tangent to the quadratic, their slopes at some point must be the same as the quadratic's and hence can be given by the same formula.

as you rightly said, the slope of the quadratic is given by the formula $f'(x) = 10 - 2x$, so we can set this for m in the equation of the line above. thus we have:

$y = (10 - 2x)x + (10 - 50 + 10x)$

$\Rightarrow y =- 2x^2 + 20x - 40$

since the lines touch the curve at one point, we equate the formulas to find what x's the lines touch the curve

thus we have: $-2x^2 + 20x - 40 = -16 + 10x - x^2$

I think you can take it from here

3. Originally Posted by Jhevon
we know that the lines must be of the form y = mx + b, right? since they pass through the point (5,10), we have by the point-slope form that te lines are of the form:

$y - 10 = m(x - 5)$

$\Rightarrow y = mx + (10 - 5m)$

Now, since the lines are tangent to the quadratic, their slopes at some point must be the same as the quadratic's and hence can be given by the same formula.

as you rightly said, the slope of the quadratic is given by the formula $f'(x) = 10 - 2x$, so we can set this for m in the equation of the line above. thus we have:

$y = (10 - 2x)x + (10 - 50 + 10x)$

$\Rightarrow y =- 2x^2 + 20x - 40$

since the lines touch the curve at one point, we equate the formulas to find what x's the lines touch the curve

thus we have: $-2x^2 + 20x - 40 = -16 + 10x - x^2$

I think you can take it from here
That seems to work yet I'm still stuck. Once I get both (a,f(a))'s I'm not sure where to go. I got (4,40) and (6,60) but am uncertain.

4. Originally Posted by pseizure2000
That seems to work yet I'm still stuck. Once I get both (a,f(a))'s I'm not sure where to go. I got (4,40) and (6,60) but am uncertain.
you're supposed to be finding the equation of the lines. you're not looking for points. just find the x's where the line intersects the parabola and then use the point-slope form to find the equation of the lines. the slope of each line will be given by f'(a), where x = a is where they cut the parabola