How do you find the important points to graph y = -x^2 + 3y=x2+3?

1 Answer
Jul 2, 2018

vertex (0, 3)(0,3)
xx-intercepts: (-sqrt(3), 0), (sqrt(3), 0)(3,0),(3,0)
yy-intercept: (0, 3)(0,3)

Explanation:

Given: y = -x^2 + 3y=x2+3

With the equation in the form: Ax^2 + Bx + C = 0Ax2+Bx+C=0,

the vertex is at (-B/(2A), f(-B/(2A)))(B2A,f(B2A)),

the axis of symmetry is x = -B/(2A)x=B2A

If the coefficient A < 0A<0, the vertex is a maximum

If the coefficient A > 0A>0, the vertex is a minimum

For the given equation:

-B/(2A) = 0/-2 = 0B2A=02=0

f(0) = -(0)^2 + 3 = 3f(0)=(0)2+3=3

vertex (0, 3)(0,3) is a maximum; axis of symmetry: x = 0x=0

color(blue) ("Find x-intercepts")Find x-intercepts by setting y = 0y=0:

0 = -x^2 + 30=x2+3

-3 = -x^23=x2

x^2 = 3 => x = +- sqrt(3)x2=3x=±3

xx-intercepts: (-sqrt(3), 0), (sqrt(3), 0)(3,0),(3,0)

color(red) ("Find y-intercept")Find y-intercept by setting x = 0x=0:

y = -(0)^2 + 3 => y = 3y=(0)2+3y=3

yy-intercept: (0, 3)(0,3), which is the vertex