Derive equations of all lines which are tangent to the graph of y= -7 - x^2 and passing through the point (3,0). (This point is not on the graph).

Question

I don't know how to start this.

So since the lines are tangent to the given equation, then the derivative will = 0, right?

So how do I find the correct x and y to put in the tangent line equation? I can't just plug in 3 for x in the original equation, can I?

oh wait, but since the derivative is 0 I can solve for x and plug the x i found in the derivative in the original equation. But how do i ensure it passes (3,0)? — Elsa, Nov 01 '14 at 23:06

score 0 · Accepted Answer · answered Nov 01 '14 at 23:14

Let's derive the equation of the tangent line at the point $(x_1,y_1)$.

The gradient is $\cfrac {dy}{dx}=-2x$. At $(x_1, y_1)$ this has the value $-2x_1$.

Now the line with gradient $m$ which passes through the point $(x_1,y_1)$ is best written as $y-y_1=m(x-x_1)$ [note below] $= -2x_1(x-x_1)$.

Now you have the equation of the line, you can find the condition for $(3,0)$ to lie on the line. You may be able to simplify it/modify it by using the condition that $(x_1,y_1)$ is on the curve.

Note: $\cfrac {dy}{dx}=\lim_{x\to x_1} \cfrac {y-y_1}{x-x_1}$ and this is directly related to the form of the tangent as $(x-x_1)\cfrac {dy}{dx}=y-y_1$ - the straight line which most closely approximates the curve.

Derive equations of all lines which are tangent to the graph of y= -7 - x^2 and passing through the point (3,0). (This point is not on the graph).

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