Finding Tangent Slope to Parabola Using Theorem 2 | Limit Help

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Find tangent slope to parabola using Theorem 2.
y(x) = x^2 + 2x \; \text{at} \; P(-3.3)

Theorem 2:
m = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}
m = \lim_{h \rightarrow 0} \frac{(a + h)^2 + 2(a + h) - 3}{h} = \lim_{h \rightarrow 0} \frac{a^2 + h^2 + 2ah + 2a + 2h - 3}{h}
\lim_{h \rightarrow 0} \frac{a^2 + h^2 + 2ah + 2a + 2h - 3}{h} = \lim_{h \rightarrow 0} \frac{(a + h - 1)(a + h + 3)}{h}
:rolleyes:

I have already solved the tangent line using the Tangent Line Theorem, however, I have been unable to eliminate h from the denominator in this theorem using division or numerator conjugates...

Any suggestions?
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either use a throughout or immediatedly plug in its value. Then
m=lim (h2+2ah+2h)/h=2(a+1)
 
Isn't a=-3...?

Daniel.
 
\lim_{h\rightarrow 0} \frac{y(x+h) - y(x)}{h}

y(x) = x^2 + 2x

\lim_{h\rightarrow 0} \frac{(x+h)^2 + 2(x+h) - x^2 - 2x}{h}

\lim_{h\rightarrow 0} \frac{x^2 + 2xh + h^2 + 2x + 2h - x^2 - 2x}{h}

\lim_h{\rightarrow 0} \frac{2xh + h^2 + 2h}{h}

Can you see what's going on now?
 

Solved!, thanks everyone!
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