For the graph of y = 4 + 3x − x², determine a) The coordinates of the point where the curve cuts the y-axis b) The coordinates of the point where the curve cuts the x-axis c) The coordinates of any minimum or maximum points d) Hence sketch the graph of y = 4 + 3x − x²

Substitute $x=0$ into the equation.

The coordinates where the curve cuts the y-axis are $\boxed{(0, 4)}$.

Question 1b) The coordinates of the point where the curve cuts the x-axis.

Set the equation to 0.

Rearrange the quadratic equation to standard form $ax^2 + bx + c = 0$:

We need two numbers that multiply to -4 and add to -3. These numbers are -4 and 1.

The coordinates where the curve cuts the x-axis are $\boxed{(-1, 0) \text{ and } (4, 0)}$.

Question 1c) The coordinates of any minimum or maximum points.

To add these, find a common denominator (4):

The second derivative $\frac{d^2y}{dx^2} = \frac{d}{dx}(3-2x) = -2$. Since $\frac{d^2y}{dx^2} < 0$, the point is a maximum.
The coordinates of the maximum point are $\boxed{\left(\frac{3}{2}, \frac{25}{4}\right) \text{ or } (1.5, 6.25)}$.

Question 1d) Hence sketch the graph of $y = 4 + 3x - x^2$.

Using the points calculated:

• y-intercept: $(0, 4)$
• x-intercepts: $(-1, 0)$ and $(4, 0)$
• Maximum point: $(1.5, 6.25)$

The graph is a parabola opening downwards, passing through these points.

      ^ y
      |
6.25  . (1.5, 6.25) Max
      |
4     . (0, 4)
      |
      |
      |
-1 ---.-------.-------.-----> x
      |       0       4
      |

(A textual representation of the sketch is provided above. For a precise visual sketch, plot these points on a coordinate plane and draw a smooth parabolic curve through them, opening downwards.)