How do you find the vertex and intercepts for $f(x) = 3x^2 + 6x + 1$ ?

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Answer 1 · 2017-10-13T21:08:25

Vertex (-1,-2)
y-intercept (0,1)
x-intercept 1.8165, 0.1835

$y=a*(x-h)^2+k$ is the equation of parabola, with vertex (h,k).
$y=3x^2+6x+1=3*(x^2+2x+1)-2$
$:.y=3*(x+1)^2-2$
Vertex (h,k) = $(-1,-2)$

To find y-intercept, let $x=0$
$y=3-2=1$ ;

To find x-intercept, let $y=0$
$0=3(x+1)^2-2$
$(x+1)^2=2/3$
$x=1-(+-sqrt(2/3))$
$x=1+sqrt(2/3), 1-sqrt(2/3) = 1.8165, 0.1835$

How do you find the vertex and intercepts for f(x) = 3x^2 + 6x + 1f(x) = 3x^2 + 6x + 1?