Answer:
Part A)
The slope is two.
Part B)
[tex]\displaystyle y = 2x - 7[/tex]
Step-by-step explanation:
Part A)
We want to find the slope of the curve:
[tex]\displaystyle y = x^2 - 2x - 3[/tex]
At the point P(2, -3) by using the limit of the secant slopes through point P.
To find the limit of the secant slopes, we can use the difference quotient. Recall that:
[tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}[/tex]
Since we want to find the slope of the curve at P(2, -3), x = 2.
Substitute:
[tex]\displaystyle f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}[/tex]
Simplify. Note that f(2) = -3. Hence:
[tex]\displaystyle \begin{aligned} f'(2) &= \lim_{h\to 0} \frac{\left[(2+h)^2 - 2(2+h) - 3\right] - \left[-3\right]}{h} \\ \\ &=\lim_{h \to 0}\frac{(4 + 4h + h^2)+(-4-2h)+(0)}{h} \\ \\ &= \lim_{h\to 0} \frac{h^2+2h}{h}\\ \\&=\lim_{h\to 0} h + 2 \\ \\ &= (0) + 2 \\ &= 2\end{aligned}[/tex]
(Note: I evaluated the limit using direct substitution.)
Hence, the slope of the curve at the point P(2, -3) is two.
Part B)
Since the slope of the curve at point P is two, the slope of the tangent line is also two.
And since we know it passes through the point (2, -3), we can consider using the point-slope form:
[tex]\displaystyle y - y_1 = m(x-x_1)[/tex]
Substitute. m = 2. Therefore, our equation is:
[tex]\displaystyle y + 3 = 2(x-2)[/tex]
We can rewrite this into slope-intercept if desired:
[tex]\displaystyle y = 2x - 7[/tex]
We can verify this by graphing. This is shown below:
