Question

unit 3: lesson 1 homework a
directions: find the indicated part of the parabola. you must show all work.

1. find the vertex: y = x² + 10x + 22
2. find the vertex: y = -2x² + 4x + 5
3. find the axis of symmetry: y = -x² - 6x - 5
4. find the axis of symmetry: y = 3x² - 12x + 5
5. find the minimum or maximum value: y = 2x² + 8x + 3
6. find the minimum or maximum value: y = -3x² + 7
7. find the y - intercept: y = x² - 9x + 10
8. find the y - intercept: y = x² + 16x

Explanation:

1. For \(y = x^{2}+10x + 22\)

Step1: Find x - coordinate of vertex

For a quadratic function \(y=ax^{2}+bx + c\), the x - coordinate of the vertex is \(x=-\frac{b}{2a}\). Here \(a = 1\), \(b = 10\), so \(x=-\frac{10}{2\times1}=- 5\).

Step2: Find y - coordinate of vertex

Substitute \(x=-5\) into \(y=x^{2}+10x + 22\), we get \(y=(-5)^{2}+10\times(-5)+22=25 - 50+22=-3\).

Step1: Find x - coordinate of vertex

For \(y = ax^{2}+bx + c\) with \(a=-2\), \(b = 4\), \(x=-\frac{b}{2a}=-\frac{4}{2\times(-2)} = 1\).

Step2: Find y - coordinate of vertex

Substitute \(x = 1\) into \(y=-2x^{2}+4x + 5\), we have \(y=-2\times1^{2}+4\times1+5=-2 + 4+5=7\).

Step1: Recall axis - of - symmetry formula

The axis of symmetry of a quadratic function \(y=ax^{2}+bx + c\) is \(x=-\frac{b}{2a}\). Here \(a=-1\), \(b=-6\), so \(x=-\frac{-6}{2\times(-1)}=-3\).

Answer:

The vertex is \((-5,-3)\)

2. For \(y=-2x^{2}+4x + 5\)