Science:Math Exam Resources/Courses/MATH110/December 2017/Question 07 (a)

MATH110 December 2017

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[hide] Question 07 (a)
(a) Sketch the graph of the function $${\begin{aligned}h(x)={\begin{cases}x^{2}-1&{\text{ if }}x\leq -1\\x+1&{\text{ if }}-1<x<0\\e^{x}&{\text{ if }}x\geq 0.\end{cases}}\end{aligned}}$$ Make sure you identify all $x$- and $y$-intercepts (if they exist) by writing down the coordinates of those points on your graph.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Draw each graph and restrict the graphs to the one on the assigned regions.

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[show] Solution

The graph of $y=x^{2}-1$ can be obtained by translating the graph of $y=x^{2}$ vertically by $-1$. In other words, it is the parabola that opens upward with minimum function value $-1$. Also, it has $x$-intercepts at $x=-1$ and $x=1$, which easily follows from $y=x^{2}-1=(x+1)(x-1)=0$. On the other hand, the graph of $y=x+1$ is a straight line with the slope $1$. It passes through $(-1,0)$) ($x$-intercept) and $(0,1)$ ($y$-intercept). Finally, the graph of the exponential function $y=e^{x}$ passes through $(0,1)=(0,e^{0})$ and approaches to the positive infinity ($+\infty$) as $x$ goes to the positive infinity. Combining these information and restricting them on the assigned domain, we have the following graph of $h$. Apparently in the graph, the only $x$-intercept of $h$ is $x=-1$, indicated by the red dot, while the only $y$-intercept of $h$ is $y=1$, indicated by the green dot.