Question

5、 determine the x-intercepts and indicated property of the following relationship! (2 mark)
y = 2(x + 3)(x - 5)
x-intercepts:
will this graph be wider or narrower than y = 1x²?

1. determine the vertex and indicated property of the following relationship: (2mark)

a) y = 3(x - 4)² + 2
vertex:
this parabola will have a ______________ (maximum or minimum)

1. determine the y-intercept of the parabola below: (1 mark)

y = 2(x + 4)² - 3
y-intercept:

Explanation:

Question 5

Step1: Find x-intercepts

To find x-intercepts, set \( y = 0 \). So, \( 0 = 2(x + 3)(x - 5) \). Since a product is zero when at least one factor is zero, we have \( x + 3 = 0 \) or \( x - 5 = 0 \). Solving these, \( x = -3 \) or \( x = 5 \).

Step2: Compare width

The standard parabola is \( y = x^2 \) (coefficient \( a = 1 \)). For \( y = 2(x + 3)(x - 5) \), when expanded, the coefficient of \( x^2 \) is \( 2 \). If \( |a| > 1 \), the graph is narrower; if \( |a| < 1 \), it's wider. Here, \( |2| > 1 \), so it's narrower.

Step1: Find vertex

The equation \( y = 3(x - 4)^2 + 2 \) is in vertex form \( y = a(x - h)^2 + k \), where vertex is \( (h, k) \). So, \( h = 4 \), \( k = 2 \), vertex is \( (4, 2) \).

Step2: Determine max/min

Since \( a = 3 > 0 \), the parabola opens upward, so it has a minimum.

Step1: Find y-intercept

To find y-intercept, set \( x = 0 \). Substitute \( x = 0 \) into \( y = 2(x + 4)^2 - 3 \). First, calculate \( (0 + 4)^2 = 16 \). Then, \( y = 2(16) - 3 = 32 - 3 = 29 \).

Answer:

x-intercepts: \( -3 \) and \( 5 \); Narrower

Question 6a