Question

on this slide you will need to fill out the table and graph out a parabola on the coordinate plane provided. any work you need to do can be done below in the space provided.

1. the points at which a quadratic equation intersects the x-axis are referred to as:

graph the quadratic equation and identify the solution(s).

1. $y = x^2 + 2x - 3$

solutions:

1. $y = x^2 - 8x + 12$

solutions:

1. $y = x^2 - 4$

solutions:

1. $y = -x^2 + 10x - 21$

solutions:

1. $y = x^2 - 4x + 5$

solutions:

1. $y = -2x^2 - 8x$

solutions:

do your work here. provide the number for the problem above so i can identify which work corresponds to which problem:

Explanation:

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Problem 1

Step1: Define x-axis intersection points

The points where a quadratic equation intersects the x-axis are called x-intercepts, roots, or solutions of the quadratic equation. These are the values of $x$ when $y=0$.

Answer:

(Problem 1):
x-intercepts (or roots/solutions)
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Problem 2: $y = x^2 + 2x - 3$

Step1: Complete the table

Calculate $y$ for given $x$ values:

• $x=-1$: $y=(-1)^2 + 2(-1) -3 = 1-2-3=-4$
• $x=0$: $y=0+0-3=-3$
• $x=1$: $y=1+2-3=0$
• $x=-2$: $y=(-2)^2 +2(-2)-3=4-4-3=-3$
• $x=-3$: $y=(-3)^2 +2(-3)-3=9-6-3=0$

$x$	$y$
-2	-3
-1	-4
0	-3
1	0

Step2: Find solutions (x-intercepts)

Set $y=0$: $x^2+2x-3=0$
Factor: $(x+3)(x-1)=0$
Solve: $x=-3, x=1$