Question

(a) find the slope, m, of the tangent to the curve y = 2 + 5x^2 - 2x^3 at the point where x = a.

m=

(b) find equations of the tangent lines at the following points.

(1, 5)

y=

(2, 6)

y=

(c) graph the curve and both tangents on a common axis.

Explanation:

Step1: Differentiate the function

The derivative of $y = 2+5x^{2}-2x^{3}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $y'=\frac{d}{dx}(2)+\frac{d}{dx}(5x^{2})-\frac{d}{dx}(2x^{3})$. Since $\frac{d}{dx}(2) = 0$, $\frac{d}{dx}(5x^{2})=10x$ and $\frac{d}{dx}(2x^{3}) = 6x^{2}$, we have $y'=10x - 6x^{2}$. At $x = a$, the slope $m = 10a-6a^{2}$.

Step2: Find the slope at $(1,5)$

Substitute $x = 1$ into $y'$. When $x = 1$, $y'(1)=10\times1-6\times1^{2}=4$. Using the point - slope form of a line $y - y_{1}=m(x - x_{1})$, with $(x_{1},y_{1})=(1,5)$ and $m = 4$, we get $y - 5=4(x - 1)$. Simplifying gives $y=4x + 1$.

Step3: Find the slope at $(2,6)$

Substitute $x = 2$ into $y'$. When $x = 2$, $y'(2)=10\times2-6\times2^{2}=20 - 24=-4$. Using the point - slope form of a line $y - y_{1}=m(x - x_{1})$, with $(x_{1},y_{1})=(2,6)$ and $m=-4$, we get $y - 6=-4(x - 2)$. Simplifying gives $y=-4x+14$.

Answer:

(a) $m = 10a-6a^{2}$
(b) $(1,5)$: $y = 4x + 1$; $(2,6)$: $y=-4x + 14$
(c) Graphing is a visual task. The curve $y = 2+5x^{2}-2x^{3}$ is a cubic - like curve. The tangent line $y = 4x + 1$ has a slope of 4 and y - intercept of 1. The tangent line $y=-4x + 14$ has a slope of - 4 and y - intercept of 14. You can use graphing software (e.g., Desmos) to accurately plot the curve and the two tangent lines.