How do I find the y intercept for $P(x)= (x-1)^2(x-3)$?
I am sketching a graph of this polynomial $P(x)= (x-1)^2(x-3)$. I have the zeros, and I have made a sign diagram. Thus, I have all my values except for the $y$ intercept. How do I get it? If I expand the binomial I get an expression with no constants.
$\endgroup$ 15 Answers
$\begingroup$To get the $y$-intercept we basically set $x=0$. We thus get: $$(x-1)^2(x-3)\Rightarrow (-1)^2(-3)=-3$$
Therefore the $y$-intercept is the point $(0,-3)$.
If you graph $\mathrm P(x)=(x-1)^2(x-3)$ you will see that it meets the $y$-axis at the point $(0,-3)$:
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I hope this helps.
Best wishes, $\mathcal H$akim.
First of all: Never, ever, ever expand!
Finding the $y$-intercept is finding where the curve meets the $y$-axis. To find where two curves meet you must solve both of their equations simultaneously.
Points on the $y$-axis are of the form $(0,1)$, $(0,2)$, $(0,3)$, $(0,4)$, etc. In general the $x$-coordinate is always zero for any point on the $y$-axis. The $y$-axis has the equation $x=0$.
To find where a curve $y=\mathrm{f}(x)$ means the $y$-axis we need to solve the equations $y=\mathrm{f}(x)$ and $x=0$ simultaneously. That means we will be on the curve $y=\mathrm{f}(x)$ and on the line $x=0$.
It follows that the $y$-intercept of any graph $y=\mathrm{f}(x)$ is given by $x=0$ and $y=\mathrm{f}(0)$. In your example $\mathrm{f}(x) = (x-1)^2(x-3)$. The $y$-intercepts have $y=(0-1)^2(0-3) = -3$.
By a similar argument, the $x$-intercepts are found by solving $y=\mathrm{f}(x)$ and $y=0$ simultaneously, for the $x$-axis has the equation $y=0$. To do this we solve $\mathrm{f}(x)=0$. In your case, the $x$-intercepts are given by $(x-1)^2(x-3)=0$, i.e. $x=1$, $x=1$ and $x=3$.
The significance of the double root $x=1$ is that the graph $y=(x-1)^2(x-3)$ must be tangent to the $x$-axis when $x=1$. It will cross the $x$-axis when $x=3$ and cross the $y$-axis when $y=-3$.
You now have all of the intercepts, and you know how the curve crosses at these special points. Can you sketch a curve that meets the criteria?
Evaluate $P(0)$: That is, "plug in" $0$ everywhere you see $x$ in $P(x)$.
The value, when you evaluate, is your $y$ intercept.
$\endgroup$ 2 $\begingroup$The y-intercept corresponds to $x=0$.
$\endgroup$ $\begingroup$the $y-intercept$ is where the functions $P(x)$ crosses the the $y-axis$ and this occurs obviously when $x =0 $. Hence, $P(0) = (0-1)^2(0-3) = 1 (-3) = -3 $ is the y intercept.
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