slope of Tangent line

[Hellpop80]

Consider the parabola y=2x^2. Find the slope of the tangent line at the point (5,25). See photo below.

The problem I have with this is that point (5,25) does not lie on parabola y=2x^2.

y'=4x = y'=4(5)=20 , slope=20 , 25=20(5)+b = y=20x-75 Correct?

Only the parabola and tangent line do not touch if you graph them. Am I missing something, or is this a possible typo? Thanks.

 

[MarkFL]

Yes, we would have:

[MATH]y(5)=2(5)^2=50[/MATH]
So, the point \((5,50)\) is on the parabola instead, and the tangent line would be:

[MATH]y=20x-50[/MATH]

 

[Harry_the_cat]

A typo in the question, I'd say.

 

[HallsofIvy]

Not necessarily a typo. The point (5, 25) is not on the parabola but there exist a line through (5, 25) that is tangent to the parabola at some other point. Any line through (5, 25) is of the form y= m(x- 5)+ 25 for some slope, m. That will be tangent to the parabola \(\displaystyle y= 2x^2\) where \(\displaystyle 4x= m\) and \(\displaystyle m(x- 5)+ 25= 2x^2\). So we have two equations to solve for m and x.

Since 4x= m, \(\displaystyle 4x(x- 5)+ 25= 4x^2- 20x+ 25= 4x^2\). That is, 20x= 25 so x= 5/4. Then m= 4x= 4(5/4)= 5. The line y= 4(x- 5)+ 25= 4x+ 5 passes through the point (5 25) and is tangent to the parabola \(\displaystyle y= 2x^2\) at \(\displaystyle \left(\frac{5}{4}, \frac{25}{8}\right)\).