What is the axis of symmetry and vertex for the graph $f (x) = - 4 x^{2}$ $f(x)=-4x^2$ ?

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What is the axis of symmetry and vertex for the graph $f (x) = - 4 x^{2}$ $f(x)=-4x^2$ ?

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Answer 1 · 2018-06-02T02:27:31

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Explanation:

The axis of symmetry can be calculated for a quadratic in standard form ( $a x^{2} + b x + c$ $ax^2+bx+c$ ) by the equation $x = - \frac{b}{2 a}$ $x=-b/(2a)$
In the equation in your question, $a = - 4, b = 0$ $a=-4, b=0$ , and $c = 0$ $c=0$ . Thus, the axis of symmetry is at $x = 0$ $x=0$ :
$x = - \frac{b}{2 a} = - \frac{0}{2 \cdot - 4} = \frac{0}{- 8} = 0$ $x=-b/(2a)=-0/(2*-4)=0/-8=0$

To find the vertex, substitute the x-coordinate of the axis of symmetry for x in the original equation to find its y-coordinate:
$y = - 4 x^{2} = - 4 \cdot 0^{2} = - 4 \cdot 0 = 0$ $y=-4x^2 = -4*0^2=-4*0=0$

So the axis of symmetry is $x = 0$ $x=0$ and the vertex is at $(0, 0)$ $(0,0)$ .

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What is the axis of symmetry and vertex for the graph $f (x) = - 4 x^{2}$ $f(x)=-4x^2$ ?

1 Answer

Explanation:

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