I'm having trouble understanding the right steps to do the problem:
Find all points on the graph of the function $f(x)=x^2 + 4$ that have a tangent which passes through the point $(0,0)$.
[Math] Find all points on the graph $f(x)=x^2 + 4$ of that have a tangent which passes through the point $(0,0)$
calculusderivatives
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Best Answer
Differentiating:
$f'(x) = 2x$. So the slope of the tangent line at $x_0$ is given by $m = 2x_0$.
The equation of line tangent to the curve when $x = x_0$ is given by $$y - (x_0^2 + 4) = 2x_0(x - x_0)\iff y = 2x_0x + \underbrace{(x_0^2 + 4 - 2x_0^2)}_{y\text{-intercept}\, = \,0} \tag{$*$}$$
When does the y-intercept equal zero? $$4-x_0^2 = 0\iff (2-x_0)(2+x_0) = 0 \iff x_0 = 2,\;\text{or }\,x_0=-2$$
Now, find the corresponding $y= f(x_0)$ values for each of the two solutions. Then your points are given by $\left(x_0, f(x_0)\right)$