Finding Tangent Slope to Parabola Using Theorem 2 | Limit Help

[Orion1]

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

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


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?

 

[mathman]

either use a throughout or immediatedly plug in its value. Then
m=lim (h²+2ah+2h)/h=2(a+1)

 

[dextercioby]

Isn't a=-3...?

Daniel.

 

[whozum]

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

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?

 

[Orion1]

Solved!, thanks everyone!