What is the vertex of y=-x^2+6y=−x2+6?

1 Answer
Tony B
Mar 6, 2017

(0,6)(0,6)

Explanation:

Consider the standardised form of y=ax^2+bx+cy=ax2+bx+c

Written as y=a(x^2+b/ax)+cy=a(x2+bax)+c

x_("vertex")=(-1/2)xxb/a" "->" "(-1/2)xx0/(-1) = 0xvertex=(−12)×ba → (−12)×0−1=0

The y-intercept = c = 6=c=6

As there is no bxbx term in y=-x^2+6" "y=−x2+6 the axis of symmetry is the y-axis. So the vertex is at (x,y)=(0,6)(x,y)=(0,6)

As the x^2x2 term is negative then the general shape of the curve is nn∩

Tony B

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