How do you find the vertex and the intercepts for $Y=x^2+5x-6$ ?

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Answer 1 · 2017-03-09T12:30:00

intercepts: $x = -6, 1$

vertex: $(- 5/2, - 49/4)$ .

$y=x^2+5x-6$

For the x-axis intercepts, factor and set to zero:

$y =(x+6)(x-1) = 0$

$implies x = -6, 1$

For the vertex, complete the square:

$y=(x+5/2)^2 - 25/4 -6$

$=(x+5/2)^2 - 49/4$

So this quadratic has its minimum value at $(- 5/2, - 49/4)$ .

How do you find the vertex and the intercepts for Y=x^2+5x-6Y=x^2+5x-6?