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Quiz 7
1. Let $f(x)=2x^2+1$. Find $f(x+2)$.
Solution: Calculate
$$f(x+2)=2(x+2)^2+1=2(x^2+4x+4)+1=2x^2+8x+9.$$
2. Consider $2y+3x=4$. Is $(0,0)$ a solution?
Solution: The point $(0,0)$ is suggesting that $x=0$ and $y=0$. Plug these values into the equation to get the equation
$$2(0)+3(0)=4,$$
but this equation is not true, and so $(0,0)$ is not a solution of $2y+3y=4$.
3. Consider $y-2x=5$. Is $(0,5)$ a solution?
Solution: Here we are looking at $x=0$ and $y=5$. Plug these values in to get the equation
$$5-2(0)=5,$$
which is true, and so $(0,5)$ is a solution of $y-2x=5$.
4. Consider $x+y=8$. Find a solution with $x$-coordinate $10$.
Solution: To require an $x$-coordinate of $10$ means we will set $x=10$ and solve for $y$. Plug in this $x$ to get
$$10+y=8,$$
subtract $10$ to get
$$y=-2.$$
This means that $(10,-2)$ is a solution of $x+y=8$.
5. Consider $2x-8y=4$. Find a solution with $y$-coordinate $7$.
Solution: To require a $y$-coordinate of $7$ means we will set $y=7$ and solve for $x$. Plug in this $y$ to get
$$2x-8(7)=4.$$
Simplify the arithmetic to get
$$2x-56=4.$$
Add $56$ to get
$$2x=60.$$
Now divide by $2$ to get
$$x=30.$$
This shows that $(30,7)$ is a solution to $2x-8y=4$.
