How do you find the vertex and the intercepts for $f(x)=x^2-x-12$ ?

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Answer 1 · 2018-05-08T12:32:44

Vertex is at $(0.5, -12.25)$ , x intercepts are at
$(-3,0) and(4,0)$ and y intercept is at $(0,-12)$

$f(x)= y= x^2-x-12 or y= (x^2-x +1/4)- 49/4$ or

$y= (x -0.5)^2 - 12.25$ Comparing with vertex form of

equation $f(x) = a(x-h)^2+k ; (h,k)$ being vertex we find

here $h=0.5 , k=-12.25 :.$ Vertex is at $(0.5, -12.25)$ .

y intercept is found by putting $x=0$ in the equation

$y = x^2- x-12 :. y = -12$ or at $(0,-12)$

y intercept is at $(0,-12)$

x intercept is found by putting $y=0$ in the equation

$y = x^2- x-12 or x^2- x-12 =0$ or

$x^2- 4 x +3 x-12 =0 or x(x-4) +3 (x-4) =0$ or

$(x-4)(x+3) =0 :. x = -3 and x = 4$

x intercepts are at $(-3,0) and (4,0)$

graph{x^2-x-12 [-40, 40, -20, 20]} [Ans]

How do you find the vertex and the intercepts for f(x)=x^2-x-12 f(x)=x^2-x-12 ?