How do you graph $y=3x^2-6x-2$ ?

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Answer 1 · 2018-06-13T01:53:44

graph{3x^2-6x-2 [-10, 10, -5, 5]}

$x$ -intercepts: $1-sqrt15/3" "$ and $" "1+sqrt15/3$

Vertex: $(1, -5)$

Positive (U shaped)

There are 2 parts to graphing a parabola:

1) Finding the vertex

You can find the $x$ -value for the vertex with this formula:

$x= -b/(2a)$

For this equation, $b= -6$ and $a= 3$ . Then plug this $x$ back into the original equation, $y=3x^2-6x-2$ , to find the $y$ value.

2) Finding the $x$ -intercepts by plugging values for $a$ , $b$ , and $c$ into the quadratic formula

$x= (-b+-sqrt(b^2-4ac))/(2a)$

$x= (-(-6)+- sqrt((-6)^2-4(3)(-2)))/(2*3)$

And because the first term, $3x^2$ , is positive, we know that the parabola is shaped like a right-side-up cup, or a U shape.

How do you graph y=3x^2-6x-2y=3x^2-6x-2?