How do you find the vertex and the intercepts for $f(x) = x² -10x -3$ ?

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Answer 1 · 2017-10-23T11:48:59

Vertex is at $(5,-28)$ , x-intercepts are at
$(-0.29,0) and (10.29,0)$ and y-intercept is at $(0,-3)$

$f(x)= x^2-10x-3 or f(x)= (x^2-10x+25)-25-3$ or

$f(x)=(x-5)^2-28$ . Comparing with vertex form of equation

$y=a(x-h)^2+k ; (h,k)$ being vertex , $h=5 , k= -28 :.$

vertex is at $(5,-28)$ , y-intercept is found by puuting

$x=0$ in the equation $y= 0^2-10*0-3 = -3 :.$

y-intercept is $y=-3 or (0,-3)$ . x-intercepts are found by

puuting $y=0$ in the equation $0=(x-5)^2-28$ or

$(x-5)^2=28 or x-5 = +- sqrt28 or x = 5 +- sqrt28$

$:. x~~10.29 and x~~ -0.29 :.$ x-intercepts are at

$(-0.29,0) and (10.29,0)$

graph{x^2-10x-3 [-80, 80, -40, 40]} [Ans]

How do you find the vertex and the intercepts for f(x) = x² -10x -3 f(x) = x² -10x -3?