What is the axis of symmetry and vertex for the graph $y = 2 x^{2} + 4$ $y=2x^2+ 4$ ?

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Answer 1 · 2016-10-07T02:29:07

Vertex $\Rightarrow (0, 4)$ $=> (0,4)$

axis of symmetry $\Rightarrow x = 0$ $=> x=0$

Quadratic Equation in Standard Form

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

Vertex $\Rightarrow (- \frac{b}{2 a}, f (- \frac{b}{2 a}))$ $=> (-b/(2a),f(-b/(2a)))$

$x = - \frac{b}{2 a}$ $x=-b/(2a)$

$y = f (- \frac{b}{2 a})$ $y=f(-b/(2a))$

Various ways to write the original equation

$y = f (x) = 0 = 2 x^{2} + 0 x + 4 = 2 x^{2} + 4$ $y=f(x)=0=2x^2+0x+4=2x^2+4$

Values for $a, b and c$ $a, b and c$

$a = 2$ $a=2$

$b = 0$ $b=0$

$c = 4$ $c=4$

Substitute

$x = - \frac{0}{2 (2)} = 0$ $x=-0/(2(2))=0$

$y = f (x) = f (0) = 2 {(0)}^{2} + 4 = 0 + 4 = 4$ $y=f(x)=f(0)=2(0)^2+4=0+4=4$

Vertex $\Rightarrow (0, 4)$ $=> (0,4)$

When the x variable is squared the axis of symmetry uses the $x$ $x$ value form the vertex coordinates.

axis of symmetry $\Rightarrow x = 0$ $=> x=0$

What is the axis of symmetry and vertex for the graph y=2x2+4y=2x^2+ 4?