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Section 1.1 #18: Find the intercepts and graph the equation $2x-4y=8$.
Solution: Recall that the $x$-intercept is a point of the form $(a,0)$ for some number $a$. We find them by setting $y=0$ and solving for $x$. Recall that a $y$-intercept is a point of the form $(0,b)$. We find them by setting $x=0$ and solving for $y$. In this problem, to find the $x$-intercept we set $y=0$ and look at the resulting equation $$2x=8$$ which has solution $x=4$. This means that $\left( 4,0 \right)$ is the only $x$-intercept of this equation. To find the $y$-intercept we set $x=0$ and look at the resulting equation $$-4y=8,$$ which has solution $y=-2$. Therefore the $y$-intercept of this equation is $(0,-2)$. $\blacksquare$

Section 1.1 #56: The point $(0,1)$ is on a circle that has center $(-3,5)$. Find the length of the diameter of the circle.
Solution: Recall that a point on a circle of radius $r$ must be a distance $r$ from the center point. Turning that around, if we know we have a circle with a given center and a point on that circle, we can find the radius by computing the distance from the center to the point on the circle that we know.

In this problem we know the center is $(-3,5)$ and a point on the circle $(0,1)$. To find the radius we just find the distance from $(-3,5)$ to $(0,1)$: $$\mathrm{dist}((-3,5),(0,1))=\sqrt{(0-(-3))^2+(1-5)^2}=\sqrt{9+16}=\sqrt{25}=5.$$ Therefore the radius of the circle is $5$. $\blacksquare$

Section 1.2 #16: Determine whether the given relation is a function. Identify the domain and range: $\{(3,1),(5,1),(7,1)\}$.
Solution: This relation is a function -- each input (i.e. $x$-coordinate) corresponds to only one output. The domain is the set $\{3,5,7\}$ and the range is the set $\{1\}$.

Section 1.2 #22: Given that $f(x)=5x^2+4x$, find each of the following:
a.) $f(0)$
b.) $f(-1)$
c.) $f(3)$
d.) $f(t)$
e.) $f(t-1)$
Solution: For a): $$f(0)=5(0^2)+4(0) = 0+0 = 0.$$ For b): $$f(-1)= 5(-1)^2 + 4(-1) = 5(1)-4 = 1.$$ For c): $$f(3)=5(3^2)+4(3) = 5(9)+12 = 45+12 = 57.$$ For d): $$f(t)=5t^2+4t.$$ For e): note that $(t-1)^2 = (t-1)(t-1)=t^2-2t+1$ and compute $$f(t-1) = 5(t-1)^2 + 4(t-1) = 5(t^2-2t+1)+4t-4 = 5t^2-10t+5+4t-4=5t^2-6t+1.$$

Section 1.2 #28: Find $h(0)$, $h(2)$, and $h(-x)$ for $h(x)=x+\sqrt{x^2-1}$.
Solution: Compute $$h(0)=0+\sqrt{0^2-1} = 0+\sqrt{0-1} = 0+\sqrt{-1},$$ which is not "allowed" (read page 23 carefully, it says that we will only accept real-valued outputs for functions right now). So we say that $h(0)$ is undefined.

Now compute $$h(2)=2+\sqrt{2^2-1} = 2+\sqrt{3},$$ and $$h(-x)=-x + \sqrt{(-x)^2-1} = -x + \sqrt{x^2-1}.$$