Question

1. draw a parabola with the following properties:

range of y ≥ -3
x-intercepts of -2 and 4

1. given the quadratic equation: $y = -3.2x^2 + 5.1x - 6.2$ (2 marks)

a) how do you know that the graph of the equation will be a parabola?
b) will this parabola open up or open down?
c) will this parabola be narrower than $y = 1x^2$ or wider than $y = 1x^2$?
d) what is the $y$-intercept?

Explanation:

Part 3: Drawing the Parabola

Step1: Find the vertex's x - coordinate

The x - intercepts are - 2 and 4. The x - coordinate of the vertex (axis of symmetry) is the mid - point of the x - intercepts. The formula for the mid - point of two numbers \(x_1\) and \(x_2\) is \(x=\frac{x_1 + x_2}{2}\). So, \(x=\frac{-2 + 4}{2}=\frac{2}{2}=1\).

Step2: Determine the vertex's y - coordinate

The range is \(y\geq - 3\), so the minimum value of \(y\) is - 3. Since the parabola opens upwards (because the range has a minimum value), the vertex is at \((1,-3)\).

Step3: Plot the x - intercepts and vertex

Plot the points \((-2,0)\), \((4,0)\) and \((1, - 3)\). Then, draw a parabola that passes through these points, opening upwards.

Part 4: Analyzing the Quadratic Equation \(y=-3.2x^{2}+5.1x - 6.2\)

4a)

Step1: Recall the definition of a quadratic function

A quadratic function is of the form \(y = ax^{2}+bx + c\), where \(a
eq0\). In the given equation \(y=-3.2x^{2}+5.1x - 6.2\), the highest power of \(x\) is 2 (since \(a=-3.2
eq0\)) and it is a polynomial function. The graph of a quadratic function is a parabola.

Step1: Recall the rule for the direction of the parabola

For a quadratic function \(y = ax^{2}+bx + c\), if \(a>0\), the parabola opens up; if \(a < 0\), the parabola opens down. In the equation \(y=-3.2x^{2}+5.1x - 6.2\), \(a=-3.2\), and \(-3.2<0\).

Answer:

Because the equation is in the form \(y = ax^{2}+bx + c\) (\(a=-3.2
eq0\)), which is the standard form of a quadratic function, and the graph of a quadratic function is a parabola.

4b)