2
$\begingroup$

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?

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

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

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

$\endgroup$
  • 1
    $\begingroup$ Aah, so basically the equation for the tangent should've been : $y - y_0 = 2x_0 (x - x_0)$, right? $\endgroup$ – Somenath Sinha Apr 19 '18 at 5:15
  • 1
    $\begingroup$ yes my friend.. $\endgroup$ – James Apr 19 '18 at 5:16
1
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $$ y=4x-4$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.