Find the points on the graph of $y = x^{2}+2x+6 $at which the slope of the tangent line is equal to $4$
Just starting to learn calculus. This is what I have done so far. Take the Derivative $f '(x) = 2x+2$. So if I plug in $1$ for $x$, I get $4$, so my first point is $(1,y)$. How do I get the Y coordinate?
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$\begingroup$Since you know that the $x$ coordinate giving the right slope is $1$the point on the graph will be $(1,f(1))=(1,9)$.
$\endgroup$ $\begingroup$Now that we have $x = 1$, we can solve for $y$ given our initial function: $$y = f(x) = x^2 + 2x + 6$$ $$\implies f(1) = 1^2 + 2(1) + 6 = 9.$$ Thus, when $x = 1,$ $y = 9.$
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