Question

exercise #2: one of the following graphs shows a relationship where $y$ is a function of $x$ and one does not.
(a) draw the vertical line whose equation is $x=3$ on both graphs.
(b) give all output values for each graph at an input of 3
relationship a:
relationship b:
(c) explain which of these relationships is a function and why.

exercise #3: the graph of the function $y=x^2 - 4x + 1$ is shown below.
(a) state this functions $y$-intercept.
(b) between what two consecutive integers does the larger $x$-intercept lie?
(c) draw the horizontal line $y=-2$ on this graph.
(d) using these two graphs, find all values of $x$ that solve the equation below:
$x^2 - 4x + 1=-2$
(e) verify that these values of $x$ are solutions by using store on your graphing calculator.

common core algebra ii, unit #2 - functions as cornerstones of algebra ii - lesson #1
emathinstruction, red hook, ny 12571, © 2015

Explanation:

Exercise #2

Step1: Draw $x=3$ on both graphs

(Visual: A vertical line passing through $x=3$ on both grids; for Relationship A, it crosses the ellipse at two points; for Relationship B, it crosses the parabola at one point.)

Step2: Find outputs at $x=3$

For Relationship A: The vertical line $x=3$ intersects the ellipse at $y=4$ and $y=-4$.
For Relationship B: The vertical line $x=3$ intersects the parabola at $y=4$.

Step3: Apply vertical line test

A relation is a function if every $x$-value has exactly one $y$-value. Relationship A fails (one $x$ has two $y$s), Relationship B passes.

Step1: Find $y$-intercept

The $y$-intercept occurs at $x=0$. Substitute $x=0$ into $y=x^2-4x+1$:
$y=(0)^2-4(0)+1=1$

Step2: Estimate larger $x$-intercept

The $x$-intercepts are where $y=0$. Solve $x^2-4x+1=0$ using the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1, b=-4, c=1$:
$x=\frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}$
$\sqrt{3}\approx1.732$, so the larger intercept is $2+1.732\approx3.732$, which lies between 3 and 4.

Step3: Draw $y=-2$ on the graph

(Visual: A horizontal line passing through $y=-2$ on the parabola's grid, intersecting the parabola at two points.)

Step4: Solve $x^2-4x+1=-2$

Rewrite the equation:
$x^2-4x+3=0$
Factor:
$(x-1)(x-3)=0$
Set each factor to 0:
$x-1=0$ or $x-3=0$, so $x=1$ and $x=3$

Step5: Verify solutions

For $x=1$: $y=(1)^2-4(1)+1=1-4+1=-2$, which matches $y=-2$.
For $x=3$: $y=(3)^2-4(3)+1=9-12+1=-2$, which matches $y=-2$.

Answer:

(a) (Vertical line $x=3$ drawn on both graphs as described)
(b) Relationship A: $y=4$ and $y=-4$; Relationship B: $y=4$
(c) Relationship B is a function. It passes the vertical line test: every input $x$ has exactly one output $y$, while Relationship A has an $x$-value ($x=3$) that maps to two different $y$-values, violating the function definition.

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Exercise #3