Find the slope of the curve y = x^2 - 5x - 4 at the point P(3, -10) by finding the limit of the secant slopes through point P.
Accepted Solution
A:
Answer:The slope of the curve [tex]f(x)=x^2-5x-4[/tex] at the point P(3, -10) is m = 1Step-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