How do you find the vertex and the intercepts for $f(x)= 3x^2 + 12x + 1$ ?

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Answer 1 · 2017-05-24T00:45:16

Vertex $(-2, -11)$

$f(x) = 3x^2 + 12x + 1$

x-coordinate of vertex:
$x = -b/(2a) = -12/6 = - 2$

y-coordinate of vertex: use $x=-2$
$f(-2) = 12 - 24 + 1 = - 11$

Vertex $(-2, -11)$ .

To find $x$ -intercepts, make $f(x) = 0,$ and solve the quadratic equation:

$3x^2 + 12x + 1 = 0$

$D = d^2 = b^2 - 4ac = 144 - 12 = 132 rarr$

$d = +- 2sqrt33$

There are 2 real roots (two $x$ -intercepts):

$x = - 12/6 +- (2sqrt33)/6 = - 2 +- sqrt33/3$

How do you find the vertex and the intercepts for f(x)= 3x^2 + 12x + 1 f(x)= 3x^2 + 12x + 1?