# How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)?

This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation:

\begin{align} \text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\ \frac{dy}{dx} &= 2x^{2-1} \\ &= 2x = \text{Slope at P.} \end{align}

Now, the equation for any straight line is also satisfied for the tangent:

\begin{align} y - y_0 &= m(x - x_0) \\ \implies y - y_0 &= 2x (x - x_0) \\ \text{For point P, } x_0 &= -2 \text{ and } y_0 = 4 \\ \implies y - 4 &= 2x(x+2)\\ \implies y - 4 &= 2x^2 + 4x\\ \implies y &= 2x^2 + 4x +4\\ \end{align}

This is where the problem occurs. If I were to try to solve for $y$ using: $$y = ax^2+bx+c \implies y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

I'd get:

\begin{align} y &= 2x^2+4x+4 \text{ and, at x-intercept: }\\ x &= \frac{-4 \pm \sqrt{4^2 - (4\times2\times4)}}{2\times2} \\ x &= \frac{-4 \pm \sqrt{16 - 32}}{4} \\ x &= \frac{-4 \pm 4i}{4} \\ x &= -1 \pm i \end{align}

Is this the correct direction, or did I do something wrong?

• $m$ here should be the slope at $(-2,4)$ which is $f'(-2) = 2\cdot -2 = -4$ and not '$2x$' – Guido A. Apr 19 '18 at 5:07

Notice in your first lines of reasoning at the point $P(-2,4)$ that is when $x=-2$, the slope of the tangent line at the point is

$$\frac{d}{dx} (x^2) \bigg|_{x=-2} = 2x \bigg|_{x=-2} = -4$$

Thus, we have $m=-4$ and given the point one has

$$y - 4 = -4(x+2)$$

• Aah, so basically the equation for the tangent should've been : $y - y_0 = 2x_0 (x - x_0)$, right? – Somenath Sinha Apr 19 '18 at 5:15
• yes my friend.. – James Apr 19 '18 at 5:16

The derivative of the function $f(x)$ is the slope of the tangent line at that particular value of $x$. Meaning when you differentiate $f(x)=x^2$, the tangent line at that value is at $x=-2$ so$$f'(2)=-4$$Therefore you're tangent line is actually

$$y-4=-4(x+2)$$

The others did point out your error, so I will just add the way I'd do it:
The tangent line we are looking for is in the form of $$g(x)=ax+b$$ for the function $$f(x)=x^2$$ at $x=-2$. We know that their derivate and their value most be equal at the given point, so we have that $$a=2*(-2)=-4$$ and $$(-2)^2=-4(-2)+b$$ $$4=8+b$$ $$b=-4$$ So the eqution for the tangent line is $$y=-4x-4$$ I like this method because I do not need to remember to the equation of the line through a given point.

You need a point and a slope. The point is $(2,4)$ and the slope is $y'(2)=4.$

Thus the equation is $$y-4=4(x-2)$$ or $$y=4x-4$$