Respuesta :
Answer:
The slope of the curve [tex]f(x)=x^2-5x-4[/tex] at the point P(3, -10) is m = 1
Step-by-step explanation:
The slope of the secant line is computed via the difference quotient:
[tex]\frac{f(x+h)-f(x)}{h}[/tex]
and the slope of the tangent line is computed via the derivative:
[tex]m=f(x)'= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]
We know that [tex]f(x)=x^2-5x-4[/tex] and [tex]P(3,-10)[/tex], so you need to:
- Find the slope of the secant line, using x = 3
[tex]\frac{f(x+h)-f(x)}{h}= \frac{((x+h)^2-5(x+h)-4)-(x^2-5x-4)}{h}\\=\frac{((3+h)^2-5(3+h)-4)-(3^2-5(3)-4)}{h}[/tex]
[tex]= \frac{((3+h)^2-5(3+h)-4)+10}{h}\\\mathrm{Expand}\:\left(3+h\right)^2-5\left(3+h\right)-4[/tex]
[tex]\left(3+h\right)^2 = 9+6h+h^2\\-5\left(3+h\right) = -15-5h[/tex]
[tex]=\frac{9+6h+h^2 -15-5h+10}{h}[/tex]
[tex]\mathrm{Simplify}\:9+6h+h^2-15-5h-4+10 = h^2+h[/tex]
[tex]= \frac{h^2+h}{h}[/tex]
[tex]\mathrm{Factor}\:h^2+h = h\left(h+1\right)[/tex]
[tex]= \frac{h(h+1)}{h}[/tex]
[tex]=h+1[/tex]
2. Find the slope of the tangent
[tex]m=f(x)'= \lim_{h \to 0} h+1[/tex]
[tex]m=f(x)'= \lim_{h \to 0} h+1 = 1[/tex]
Therefore the slope of the curve [tex]f(x)=x^2-5x-4[/tex] at the point P(3, -10) is m = 1
The slope of the secant curve is 1.
Given
The given curve is;
[tex]\rm y = x^2 - 5x - 4[/tex]
At point P(3, -10).
What is the secant slope?
A secant line intersects at 2 or more points and has a slope equal to the average rate of change between those points.
The slope of the curve is calculated with the points (3+h, f(3+h)).
[tex]\rm Slope = \dfrac{f(3+h)-(-10)}{3+h-3}\\\\Slope =\dfrac{ (3+h)^2-5(3+h)-4+10}{h}\\\\Slope=\dfrac{9+h^2+6h-5h-15+6}{h}\\\\Slope=\dfrac{ h^2+h}{h}\\\\Slope = \dfrac{h(h+1)}{h}\\\\Slope = h+1[/tex]
When the value of h is 0.
Then,
The value of slope is
Slope = h + 1 = 0 + 1 = 1
Hence, the slope of the secant curve is 1.
To know more about Secant slope click the link given below.
https://brainly.com/question/17422299
