I have a small question, the task is finding an equation for the tangent line at $g(x)=2x^3+5x^2$ when $x=-2$.
My question is, instead of using calculus, what stops us from just plugging in a $x=3$ and go old fashion $y-y_1=m(x-x_1)$? Since we have the formula, why would we need to further complicate it? Or am I missing something? Thanks in advance!
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One way to get an equation of a line is with a point and a slope. This is the appropriate way with the equation you give: $y-y_1 = m(x-x_1)$. You indicate the easy way to get the point yourself, plug in $x=3$. All that remains is the slope, $m$. This is where calculus kicks in. Easiest way to get the slope at $x=3$ is to take the derivative and plug in 3. $$g'(x)=6x^2+10x$$ $$g'(3)=84$$ So $m=84$ at $x=3$.
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You could ask for the equation of a line through the point $(-2,4)$ which intersects $g(x)$ with multiplicity greater than $1$. For the line we take the equation to be $y-4=m(x+2)$, so $y=mx+2m+4$. Then \begin{align*} 2x^3+5x^2 &= mx+2m+4 \text{ so}\\ 2x^3+5x^2 - mx - 2m- 4 &= 0 \end{align*} But we know $x=-2$ must be a root of the cubic, and after dividing by $x+2$ we get the quadratic $$2x^2+x-(m+2)$$ If the intersection multiplicity is to be greater than $1$, the quadratic also needs to be divisible by $x+2$ The remainder after performing polynomial long division is $-m-2+6$. For the remainder to be zero as we want, we must have $m=4$. That gives $y=4x+12$ as the equation of the line.
Comparing what we would find using calculus, $g'(x)=6x^2+10$ giving $g'(-2)= 4$ for the slope of the tangent line at $x=-2$ which actually seems a bit less complicated, no?
You certainly can't do this.
Yes, the slope of line can be found with two points, but you can't go tossing random numbers into a function to get a tangent line. The reason why we need calculus is the notion of limit: we need two points close to each other so that they're pretty much the same point - hence giving a tangent line.