How do you find the intercepts for $y=x^2-9x+20$ ?

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Answer 1 · 2016-02-13T18:05:21

You set $y=0$ for the $x$ intercepts and $x=0$ for the $y$ intercepts. Then you solve the equations.

$x=5$
$x=4$
$y=20$

y intercepts

$y=0$

$x^2-9x+20=0$

$Δ=b^2-4ac$

$Δ=(-9)^2-4*1*20=1$

$x_(1,2)=(-b+-sqrt(Δ))/(2a)$

$x_(1,2)=(-(-9)+-sqrt(1))/(2*1)$

$x_(1,2)=(9+-1)/2$

$x_1=5$ 1st y intercept

$x_2=4$ 2nd y intercept

graph{x^2-9x+20 [-6.84, 15.67, -2, 9.27]}

x intercepts

$x=0$

$y=0^2-9*0+20$

$y=20$ 1st (and only) x intercept

graph{x^2-9x+20 [-28.64, 29.12, -0.73, 28.17]}

How do you find the intercepts for y=x^2-9x+20y=x^2-9x+20?