Question

name date intervals of increase: minimum: intervals of decrease: domain: maximum: range: set use the given graphs to find the vertex form of the equation. use the equation to find the exact ( x )-intercepts. 3. vertex form of equation: ( x )-intercepts: unit 3, lesson 1 – ready, set, go 2 of 6 download for free at openup

Explanation:

First Graph Analysis (Balance vs Time)

Step1: Identify increasing intervals

Look for x-values where the graph rises.
Interval: $(-3, 3) \cup (6, 7)$

Step2: Identify decreasing intervals

Look for x-values where the graph falls.
Interval: $(-2, -3) \cup (7, 8)$

Step3: Locate maximum value

Find the highest y-value on the graph.
Value: $2$

Step4: Locate minimum value

Find the lowest y-value on the graph.
Value: $-6$

Step5: Define domain

List all valid x-values (time) from the graph.
Domain: $[-2, 8]$

Step6: Define range

List all valid y-values (balance) from the graph.
Range: $[-6, 2]$

Step1: Identify vertex of parabola

The vertex (lowest point) is $(1, -4)$. Vertex form is $y=a(x-h)^2+k$, where $(h,k)$ is vertex.
Substitute $h=1, k=-4$: $y=a(x-1)^2-4$

Step2: Solve for coefficient $a$

Use the x-intercept $(0,0)$ to find $a$.
$0=a(0-1)^2-4 \implies 0=a(1)-4 \implies a=4$

Step3: Write vertex form equation

Substitute $a=4$ into the vertex form.
$y=4(x-1)^2-4$

Step4: Find x-intercepts

Set $y=0$ and solve for $x$.
$0=4(x-1)^2-4$
$\frac{0}{4}=(x-1)^2-1$
$1=(x-1)^2$
$x-1=\pm1$
$x=1+1=2$ or $x=1-1=0$

Answer:

Intervals of Increase: $(-3, 3) \cup (6, 7)$
Intervals of Decrease: $(-2, -3) \cup (7, 8)$
Maximum: $2$
Minimum: $-6$
Domain: $[-2, 8]$
Range: $[-6, 2]$

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Second Parabola Graph (Problem 3)