How can I show that a linear equation is a tangent to a quadratic curve? The question is:
Show that $y = 11x + 5$ is a tangent to curve $y = 3x^2 + 5x + 8$.
I have no clue about how to go about figuring this out. 
Should I graph both curves? Or should I use a certain formula?
 A: You are essentially dealing with a quadratic and a linear function. So in such a case, we can set them equal: $3x^2+5x+8=11x+5 $ which is $3x^2-6x+3=0$. This is a quadratic equation of which it can be verified that its discriminant is equal to zero (the discriminant is $b^2-4ac$). So this means that there is only one solution. A quadratic function and a linear function that only have one point of intersection is only possible if the line is tangent to the parabola. Of course, solving the quadratic equation easily shows that the x-coordinate is $x=1$ from which the $y$ can be found as well. No calculus needed here. The calculus of course would confirm that for $x=1$ the derivative of the parabola would yield 11 but I leave that up to you to verify
A: There are four things that can happen with a parabola and a line in the plane:


*

*They never intersect.

*They intersect exactly twice - i.e. the line is a secant to the parabola.

*They intersect exactly once and the line is tangent to the parabola.

*They intersect exactly once where the line crosses the parabola.


If the parabola is upright (i.e. it's of the form $y = ax^2 + bx + c$) then 4 can only happen if the line is vertical (i.e. it's of the form $x = m$), so since that isn't the case here you just have to prove that we are in case 3, which you do by (1) equating the line to the parabola, (2) solving for points of intersection, and (3) showing that there is exactly one of them, probably via the discriminant of the resulting equation.
A: $y=11x+5$ is in slope intercept form ($y=mx+b$ where m is the slope). The derivative of a function represents the slope of a tangent line at a point so we can set $y'=m$.$$y=3x^2+5x+8$$
$$y'=6x+5$$
$$11=6x+5$$
$$x=1$$
Now you must plug this x-value back into the original function to get a corresponding y-value and write a tangent line to compare to the one given.
$$y(1)=3(1)^2+5(1)+8=16$$
So the coordinate is (1,16)
Now write a tangent line in point-slope form:
$$(y-16)=11(x-1)$$
Which if you do some algebraic manipulations you will find it is in fact the same as $$y=11x+5$$.
