Question

Please help me with these calculus bc questions

Answer 1

4. Compute the derivative.

$y = 2x^2 - x - 1 \implies \dfrac{dy}{dx} = 4x - 1$

Find when the gradient is 7.

$4x - 1 = 7 \implies 4x = 8 \implies x = 2$

Evaluate $y$ at this point.

$y = 2\cdot2^2-2-1 = 5$

The point we want is then (2, 5).

5. The curve crosses the $x$-axis when $y=0$. We have

$y = \dfrac{x - 4}x = 1 - \dfrac4x = 0 \implies \dfrac4x = 1 \implies x = 4$

Compute the derivative.

$y = 1 - \dfrac4x \implies \dfrac{dy}{dx} = -\dfrac4{x^2}$

At the point we want, the gradient is

$\dfrac{dy}{dx}\bigg|_{x=4} = -\dfrac4{4^2} = \boxed{-\dfrac14}$

6. The curve crosses the $y$-axis when $x=0$. Compute the derivative.

$\dfrac{dy}{dx} = 3x^2 - 4x + 5$

When $x=0$, the gradient is

$\dfrac{dy}{dx}\bigg|_{x=0} = 3\cdot0^2 - 4\cdot0 + 5 = \boxed{5}$

7. Set $y=5$ and solve for $x$. The curve and line meet when

$5 = 2x^2 + 7x - 4 \implies 2x^2 + 7x - 9 = (x - 1)(2x+9) = 0 \implies x=1 \text{ or } x = -\dfrac92$

Compute the derivative (for the curve) and evaluate it at these $x$ values.

$\dfrac{dy}{dx} = 4x + 7$

$\dfrac{dy}{dx}\bigg|_{x=1} = 4\cdot1+7 = \boxed{11}$

$\dfrac{dy}{dx}\bigg|_{x=-9/2} = 4\cdot\left(-\dfrac92\right)+7=\boxed{-11}$

8. Compute the derivative.

$y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b$

The gradient is 8 when $x=2$, so

$2a\cdot2 + b = 8 \implies 4a + b = 8$

and the gradient is -10 when $x=-1$, so

$2a\cdot(-1) + b = -10 \implies -2a + b = -10$

Solve for $a$ and $b$. Eliminating $b$, we have

$(4a + b) - (-2a + b) = 8 - (-10) \implies 6a = 18 \implies \boxed{a=3}$

so that

$4\cdot3+b = 8 \implies 12 + b = 8 \implies \boxed{b = -4}$.