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

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Answer 1 · 2016-05-14T14:46:44

$Vertex \to (x, y) \to (- 2, 0)$ $color(blue)("Vertex "->(x,y)->(-2,0)$

$Axis of symmetry \to x = - 2$ $color(blue)("Axis of symmetry "->x=-2$

Consider the standard form $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

Write this as $y = a (x^{2} + \frac{b}{a} x) + c$ $y=a(x^2+b/ax)+c$

Then $x_{vertex} = axis of symmetry = (- \frac{1}{2}) \times \frac{b}{a}$ $x_("vertex") = "axis of symmetry "=(-1/2)xxb/a$

In this case $a = 1$ $a=1$

So for $y = x^{2} + 4 x + 4$ $y=x^2+4x+4$

$x_{vertex} = (- \frac{1}{2}) \times 4 = - 2$ $x_("vertex")=(-1/2)xx4=-2$

So by substitution for $x$ $x$

$y_{vertex} = {(- 2)}^{2} + 4 (- 2) + 4 = 0$ $y_("vertex")=(-2)^2+4(-2)+4" "=" "0$

$Vertex \to (x, y) \to (- 2, 0)$ $color(blue)("Vertex "->(x,y)->(-2,0)$

$Axis of symmetry \to x = - 2$ $color(blue)("Axis of symmetry "->x=-2$

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