The tangent to the curve y = ax^2 + bx at (2, | vedclass

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Question

(C) Given the curve $y = ax^2 + bx$.First,find the derivative $\frac{dy}{dx} = 2ax + b$.The slope of the tangent at $(2, -8)$ is $\left( \frac{dy}{dx}

Vedclass · Accepted Answer

$a = 2, b = -8$

Vedclass · Answer

(C) Given the curve $y = ax^2 + bx$.First,find the derivative $\frac{dy}{dx} = 2ax + b$.The slope of the tangent at $(2, -8)$ is $\left( \frac{dy}{dx} \right)_{(2, -8)} = 2a(2) + b = 4a + b$.Since the tangent is parallel to the $x$-axis,its slope must be $0$.Therefore,$4a + b = 0 \Rightarrow b = -4a$ ..... $(i)$.Since the point $(2, -8)$ lies on the curve,it must satisfy the equation $y = ax^2 + bx$.Substituting $x = 2$ and $y = -8$: $-8 = a(2)^2 + b(2) \Rightarrow -8 = 4a + 2b$ ..... $(ii)$.Substitute $b = -4a$ from $(i)$ into $(ii)$:$-8 = 4a + 2(-4a)$$-8 = 4a - 8a$$-8 = -4a \Rightarrow a = 2$.Now,find $b$ using $b = -4a$:$b = -4(2) = -8$.Thus,$a = 2$ and $b = -8$.

The tangent to the curve $y = ax^2 + bx$ at $(2, -8)$ is parallel to the $x$-axis. Then:

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