Let’s analyze the two problems from the uploaded image and solve them step by step:

Problem 5

The quadratic function is given as: $f(x) = (x + 2)(x - 4)$

1. Expand and Rewrite:

Expanding the equation: $f(x) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8$

2. Identify Key Features:

• X-Intercepts: Set $f(x) = 0$: $(x + 2)(x - 4) = 0 \quad \Rightarrow \quad x = -2 \text{ and } x = 4$ The x-intercepts are: $(-2, 0)$ and $(4, 0)$.

• Y-Intercept: Set $x = 0$: $f(0) = (0 + 2)(0 - 4) = -8$ The y-intercept is: $(0, -8)$.

• Vertex: The vertex lies at the midpoint of the x-intercepts: $x_{\text{vertex}} = \frac{-2 + 4}{2} = 1$ Substitute $x = 1$ into $f(x)$: $f(1) = (1 + 2)(1 - 4) = 3(-3) = -9$ The vertex is: $(1, -9)$.

3. Graph:

• Plot the x-intercepts $(-2, 0)$ and $(4, 0)$, the y-intercept $(0, -8)$, and the vertex $(1, -9)$.
• The parabola opens upwards as the coefficient of $x^2$ is positive.

Problem 6

The quadratic function is given as: $g(x) = (x - 2)^2 - 1$

1. Expand and Rewrite:

Expanding the equation: $g(x) = x^2 - 4x + 4 - 1 = x^2 - 4x + 3$

2. Identify Key Features:

• Y-Intercept: Set $x = 0$: $g(0) = (0 - 2)^2 - 1 = 4 - 1 = 3$ The y-intercept is: $(0, 3)$.

• Vertex: The vertex form directly gives the vertex as: $(2, -1)$.

• Another Point: For $x = 1$: $g(1) = (1 - 2)^2 - 1 = 1 - 1 = 0$ Another point is: $(1, 0)$.

• The parabola opens upwards because the coefficient of $x^2$ is positive.

3. Graph:

• Plot the vertex $(2, -1)$, the y-intercept $(0, 3)$, and the point $(1, 0)$.
• Reflect points to the right of the vertex to the left (e.g., reflect $(1, 0)$ to $(3, 0)$).

Would you like me to graph these functions for clarity? Let me know! 😊

Related Questions:

1. How can we identify whether a parabola opens upwards or downwards?
2. What are the differences between the factored form and vertex form of a quadratic function?
3. How can you determine the axis of symmetry of a parabola from its equation?
4. How do transformations like shifting and stretching affect the graph of a parabola?
5. How do we find the roots of a quadratic equation if it does not factor easily?

Tip:

The vertex form of a quadratic function is very useful for quickly identifying the vertex and transformations of the parent function $y = x^2$.