Expert: Josh - 7/31/2006

Question
hi, i was wondering if you could help me with this

i don't know how to find the equations of tangents whether it is at a point, from a point, through a point, where the tangent is horizontal. examples in my notes have shown finding the derivative of the original equation, but then i get lost after that step.

I have one question in particular:
Find equations of the tangent lines through (3, -2) to y = x^2 - 7

Any help would be great, thanks!  

Answer
Hi Amanda,

I think that you already know a great deal about tangents than you realize. Let us review a few points.

A tangent is a straight line which intersects with (touches) a curve at a single point. So, its equation has the form y=mx+b, where "m" is the all important gradient (or slope) of the tangent; and "b" is the point where the tangent crosses the y-axis.

To find the equation of the tangent, we need to determine "m" and "b". Usually, given a point (x1,y1) where the tangent touches a curve y, we can work out "m" by differentiating y=f(x) with respect to x, then evaluating dy/dx at x=x1; provided that the gradient is finite.

Once "m" is known, "b" may be evaluated by substituting a point lying on the tangent (you may also use the point where the tangent actually intersects with the curve) into the equation y=mx+b which represents the tangent.

In your question, we have a parabola y = x^2 - 7.
The general idea is still the same. We need to work out "m", the slope of the tangent. However, we are not given a point where the tangent touches the curve.

We proceed by first finding the intersection(s) between the tangent (y-yo)=m(x-xo) ...[#1] and the parabola y=x^2-7...[#2]

Substituting [#2] into [#1] gives
x^2-(7+yo)=m(x-xo).
We rearrange this into a quadratic of x, such that
x^2-mx+(m*xo-7-yo)=0 ...[#3]

Recall that a quadratic equation a*x^2 + b*x + c =0 has solutions (roots) x=[-b+sqrt(b^2-4ac)]/2a, [-b-sqrt(b^2-4ac)]/2a. A quadratic has only one root, when the quantity D=b^2-4ac, called the determinant, is equal to 0. This condition guarantees that the tangent touches the parabola at only one place.

Continuing from [#3], comparing the coefficients, we have a=1, b=-m, c=mxo-(7+yo). We need to satisfy the condition
D=b^2-4ac=0, in order to obtain a tangent.
This is equivalent to saying
 m^2-4*[mxo-(7+yo)]=0, i.e.,
 m^2-4m*xo+28+4yo = 0, since xo=3,yo=-2 is on tangent
 m^2-4m*3+28+4(-2)= 0, simplifying
 m^2-12m+20 = 0, factorizing
 (m-10)(m-2)= 0, which implies
m=2, or m=10 if the tangent were to intersect with the parabola.

If m=2, we have y=2x+b. In this case, putting (x=3,y=-2) into the equation yields b=y-2x=-2-2(3)=-8.
=> Solution 1: y=2x-8

If m=10, we have y=10x+b. In this case, putting (x=3,y=-2) into the equation yields b=y-10x=-2-10(3)=-32.
=> Solution 2: y=10x-32

I'm in a rush finishing work to head home. Ask me for clarification if anything doesn't make sense.

Josh.