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How do you find the vertex and the intercepts for $f (x) = x^{2} - 4$ $f(x)=x^2-4$ ?

1 Answer

Apr 8, 2016

vertex: $(0, - 4)$ $(0,-4)$
x-intercepts: $\pm 2$ $+-2$
y-intercept: $(- 4)$ $(-4)$

Explanation:

General vertex form
$XXX f (x) = m {(x - a)}^{2} + b$ $color(white)("XXX")f(x)=color(green)(m)(x-a)^2+b$ with vertex at $(a, b)$ $(a,b)$

Writing given equation: $f (x) = x^{2} - 4$ $f(x)=x^2-4$ in vertex form:
$XXX f (x) = 1 {(x - 0)}^{2} + (- 4)$ $color(white)("XXX")f(x)=1(x-0)^2+(-4)$ with vertex at $(0, - 4)$ $(0,-4)$

x-intercepts are the possible values of $x$ $x$ when $f (x) = 0$ $f(x)=0$
$XXX 0 = x^{2} - 4$ $color(white)("XXX")0=x^2-4$
$XXX \to x = \pm 2$ $color(white)("XXX")rarr x=+-2$

y-intercept is the value of $y (= f (x))$ $y(=f(x))$ when $x = 0$ $x=0$
$XXX y = f (x = 0) = 0^{2} - 4 = - 4$ $color(white)("XXX")y=f(x=0) = 0^2-4=-4$

graph{x^2-4 [-5.717, 6.773, -4.885, 1.355]}

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