SOLUTION: Graph the quadratic equation, and find the following y=-(x-3)^2+4 Vertex: X-intercepts (zeros): Axis of symmetry equation: y-intercept:

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Graph the quadratic equation, and find the following y=-(x-3)^2+4 Vertex: X-intercepts (zeros): Axis of symmetry equation: y-intercept:      Log On


   



Question 412883: Graph the quadratic equation, and find the following
y=-(x-3)^2+4
Vertex:
X-intercepts (zeros):
Axis of symmetry equation:
y-intercept:

Found 2 solutions by rfer, lwsshak3:
Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
vertex=(3,4)
zeros 1, 5
axis=3
y int=-5

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Graph the quadratic equation, and find the following
y=-(x-3)^2+4
Vertex:
X-intercepts (zeros):
Axis of symmetry equation:
y-intercept:
..
Standard form of a parabola:y=(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. Parabola opens upward if coefficient of (x-h)^2 term is positive and downward if the coefficient is negatve.
..
Given parabola opens downward.(by inspection)
Vertex: (3,4) (by inspection)
Axis of symmetry: x=3 (by inspection)
solving for y-intercept
let x=0
y=-9+4=-5
y-intercept=5
solving for x-intercepts
let y=0
(x-3)^2=4
x-3=+-sqrt(4)
x=3+-2
x=1
x=5
see the graph below:
..
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+-%28x-3%29%5E2%2B4%29+