4. Compute the derivative.
[tex]y = 2x^2 - x - 1 \implies \dfrac{dy}{dx} = 4x - 1[/tex]
Find when the gradient is 7.
[tex]4x - 1 = 7 \implies 4x = 8 \implies x = 2[/tex]
Evaluate [tex]y[/tex] at this point.
[tex]y = 2\cdot2^2-2-1 = 5[/tex]
The point we want is then (2, 5).
5. The curve crosses the [tex]x[/tex]-axis when [tex]y=0[/tex]. We have
[tex]y = \dfrac{x - 4}x = 1 - \dfrac4x = 0 \implies \dfrac4x = 1 \implies x = 4[/tex]
Compute the derivative.
[tex]y = 1 - \dfrac4x \implies \dfrac{dy}{dx} = -\dfrac4{x^2}[/tex]
At the point we want, the gradient is
[tex]\dfrac{dy}{dx}\bigg|_{x=4} = -\dfrac4{4^2} = \boxed{-\dfrac14}[/tex]
6. The curve crosses the [tex]y[/tex]-axis when [tex]x=0[/tex]. Compute the derivative.
[tex]\dfrac{dy}{dx} = 3x^2 - 4x + 5[/tex]
When [tex]x=0[/tex], the gradient is
[tex]\dfrac{dy}{dx}\bigg|_{x=0} = 3\cdot0^2 - 4\cdot0 + 5 = \boxed{5}[/tex]
7. Set [tex]y=5[/tex] and solve for [tex]x[/tex]. The curve and line meet when
[tex]5 = 2x^2 + 7x - 4 \implies 2x^2 + 7x - 9 = (x - 1)(2x+9) = 0 \implies x=1 \text{ or } x = -\dfrac92[/tex]
Compute the derivative (for the curve) and evaluate it at these [tex]x[/tex] values.
[tex]\dfrac{dy}{dx} = 4x + 7[/tex]
[tex]\dfrac{dy}{dx}\bigg|_{x=1} = 4\cdot1+7 = \boxed{11}[/tex]
[tex]\dfrac{dy}{dx}\bigg|_{x=-9/2} = 4\cdot\left(-\dfrac92\right)+7=\boxed{-11}[/tex]
8. Compute the derivative.
[tex]y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b[/tex]
The gradient is 8 when [tex]x=2[/tex], so
[tex]2a\cdot2 + b = 8 \implies 4a + b = 8[/tex]
and the gradient is -10 when [tex]x=-1[/tex], so
[tex]2a\cdot(-1) + b = -10 \implies -2a + b = -10[/tex]
Solve for [tex]a[/tex] and [tex]b[/tex]. Eliminating [tex]b[/tex], we have
[tex](4a + b) - (-2a + b) = 8 - (-10) \implies 6a = 18 \implies \boxed{a=3}[/tex]
so that
[tex]4\cdot3+b = 8 \implies 12 + b = 8 \implies \boxed{b = -4}[/tex].
