Sketch the graph of y = x2 − 4x + 3 and state the effect of the coefficient of x2 on the shape of a parabola.

For a parabola $y = ax^2 + bx + c$, the x-coordinate of the vertex is $x = -\frac{b}{2a}$.
Here, $a=1$, $b=-4$, $c=3$.

Substitute $x=2$ into the equation to find the y-coordinate:

The turning point is $(2, -1)$.

• y-intercept: Set $x=0$. $$y = (0)^2 - 4(0) + 3 = 3$$ The y-intercept is $(0, 3)$.
• x-intercepts: Set $y=0$. $$x^2 - 4x + 3 = 0$$ Factor the quadratic equation: $$(x-1)(x-3) = 0$$ So, $x=1$ or $x=3$. The x-intercepts are $(1, 0)$ and $(3, 0)$.

The axis of symmetry is a vertical line passing through the x-coordinate of the turning point.
The axis of symmetry is $x=2$.

Summary of features for sketch:

• Turning point: $\boxed{(2, -1)}$
• Intercepts: $\boxed{(0, 3), (1, 0), (3, 0)}$
• Axis of symmetry: $\boxed{x=2}$ (Since I cannot draw a graph, these points and the axis of symmetry define the parabola, which opens upwards because the coefficient of $x^2$ is positive.)

d) State the effect of the coefficient of $x^2$ on the shape of a parabola.
The coefficient of $x^2$ (denoted as $a$) determines the direction and width of the parabola. If $a > 0$, the parabola opens upwards. If $a < 0$, it opens downwards. The magnitude of $a$ affects the width: a larger $|a|$ results in a narrower parabola, while a smaller $|a|$ results in a wider parabola.

Question 2:

a) Solve graphically: $x^2 - 4x = 0$
To solve $x^2 - 4x = 0$ graphically, you would plot the graph of $y = x^2 - 4x$ and find the x-coordinates where the graph intersects the x-axis (i.e., where $y=0$).
Algebraically:

Factor out $x$:

Set each factor to zero:

The solutions are $\boxed{x=0, x=4}$.

b) Find the point(s) of intersection between $y = x^2 - 2x$ and $y = x + 2$

Here, $a=1$, $b=-3$, $c=-2$.

For $x_1 = \frac{3 + \sqrt{17}}{2}$:

For $x_2 = \frac{3 - \sqrt{17}}{2}$:

The points of intersection are $\boxed{\left(\frac{3 + \sqrt{17}}{2}, \frac{7 + \sqrt{17}}{2}\right)}$ and $\boxed{\left(\frac{3 - \sqrt{17}}{2}, \frac{7 - \sqrt{17}}{2}\right)}$.

Question 3:

a) Given $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 1 \\ 2 & 0 \end{pmatrix}$. Compute $A+B$ and $A-B$.

b) Find $AB$.

c) Calculate the determinant of $\begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}$.
For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

d) Given vectors $u = (2,3)$ and $v = (-1,4)$, find $u+v$ and $2u$.