Question

Please help me with these calculus bc questions

Answer 1

4. Compute the derivative.

`y = 2x^2 - x - 1 \implies (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 = \frac{x - 4}x = 1 - \frac4x = 0 \implies \frac4x = 1 \implies x = 4`

Compute the derivative.

`y = 1 - \frac4x \implies (dy)/(dx) = -\frac4{x^2}`

At the point we want, the gradient is

`(dy)/(dx)\bigg|_(x=4) = -\frac4{4^2} = \boxed{-\frac14}`

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

`(dy)/(dx) = 3x^2 - 4x + 5`

When
`x=0`, the gradient is

`(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 = -\frac92`

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

`(dy)/(dx) = 4x + 7`

`(dy)/(dx)\bigg|_(x=1) = 4\cdot1+7 = \boxed{11}`

`(dy)/(dx)\bigg|_(x=-9/2) = 4\cdot\left(-\frac92\right)+7=\boxed{-11}`

8. Compute the derivative.

`y = ax^2 + bx \implies (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}`.