How do you find the vertex and axis of symmetry of $f (x) = 3 x^{2} + 12 x + 1$ $f(x)= 3x^2 + 12x + 1$ ?

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How do you find the vertex and axis of symmetry of $f (x) = 3 x^{2} + 12 x + 1$ $f(x)= 3x^2 + 12x + 1$ ?

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Answer 1 · 2016-03-19T21:48:32

This is a quadratic equation of a parabola (the squared term gives it away)

$y = a x^{2} + b x + c$ $y = ax^2+bx+c$

the vertex is located where $x = - \frac{b}{2 a}$ $x = -b/(2a)$

this occurs where $x = - \frac{12}{2 \times 3}$ $x = -12/(2 xx 3)$ or at $x = - 2$ $x = -2$

substitute into the equation to figure out the y coordinate of the vertex.
The axis of symmetry is the vertical line passing through the vertex which is $x = - 2$ $x = -2$

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How do you find the vertex and axis of symmetry of $f (x) = 3 x^{2} + 12 x + 1$ $f(x)= 3x^2 + 12x + 1$ ?

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How do you find the vertex and axis of symmetry of $f (x) = 3 x^{2} + 12 x + 1$ $f(x)= 3x^2 + 12x + 1$ ?