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I'm learning function notation and will soon be doing Calculus - having trouble with this question:

Question: Find the x- and y- intercept of each function:

$f(x) = x^2 + 3x$

If I set x to 0, I find out that Y would be obviously equal to 0. I'm not sure how to do that for the other value.

I try:

$f(x) = x^2 + 3x$

$y = x^2 + 3x$

$0 = x^2 + 3x$

$-3x = x^2$

I don't know what to do from there, if I even did anything correct

On another topic, is their any easy introductory book for Calculus I?

Garena
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  • the $y$-intercept is when $x=0$; the $x$-intercepts are when $y=0$. you correctly found that when $x=0$, $y=0$, so the $y$-intercept is $(0,0)$. to find the $x$-intercepts, your approach is right; you just need to solve the equation $x^2+3x=0$ by factoring the $x$, so $x(x+3)=0$. hence $x=0$ or $x=-3$. so the $x$-intercepts are $(0,0)$ and $(-3,0)$. – symplectomorphic May 05 '14 at 09:32
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    $0=x^2+3x=x \times (x+3)$ so, either $x=0$ or $x+3=0$. – Claude Leibovici May 05 '14 at 09:33

1 Answers1

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Instead of subtracting $3x$, factor your right hand side. $$x^2+3x=x(x+3)=0$$ Now using the Zero Property of Multiplication, which states "if $ab=0,$ then $a=0$ or $b=0$, " $x=0$ or $x+3=0$.

Thus $x=0, x=-3$. Then your x-intercepts are $(0,0), (-3,0)$.

Eleven-Eleven
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