SOLUTION: Please help me solve the following quadratic equation. Any help or guidance you offer is greatly appreciated because I am very confused with these types of problems. Find the v

Question 23687: Please help me solve the following quadratic equation. Any help or guidance you offer is greatly appreciated because I am very confused with these types of problems.
Find the vertex and intercepts for each quadratic function, and graph.
y = -x^2 + 2x + 3
Thank you
Found 2 solutions by Earlsdon, venugopalramana:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Find the vertex:
The x-coordinate of the vertex of a parabola is given by:
Your quadratic equation is:
where: a = -1, b = 2, and c = 3
The x-coordinate of the vertex is:

The y-coordinate is found by substituting this value of x (1) into the quadratic equation and solving for y.

The vertex is at: (1, 4)
The (x) intercepts can be found by solving the quadratic equation for the two root (where the parabola crosses the x-axis, if the roots are real).
Set y = 0 because this is where the parabola will cross the x-axis.
Factor -1 from the left side.
Factor the parentheses.
Clear the -1 by dividing both sides by -1
Apply the zero product principle.
and/or
The roots (x-intercepts) are:
x = 3
x = -1
Here's the graph:

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find the vertex and intercepts for each quadratic function, and graph.
y = -x^2 + 2x + 3
WRITE AS A SQUARE USING X^2 AND X TERMS
Y=3-{(X^2-2*X*1+1^2)-1^2}
Y=3-(X-1)^2+1=4-(X-1)^2
SINCE (X-1)^2 IS A PERFECT SQUARE IT IS ALWAYS POSITIVE OR MINIMUM ZERO.SO WE GET A MAXIMUM VALUE FOR Y AT X=1..VALUE IS Y=4..THIS IS CALLED VERTEX.YOU CAN SEE IT IN THE GRAPH.
INTERCEPTS MEAN WHERE THE CURVE CUTS THE AXES
PUT X=0 AND FIND Y TO GET Y INTERCEPTS..WE GET Y=3 HENCE 3 IS THE Y INTERCEPT.
PUT Y=0 AND FIND X TO GET X INTERCEPTS...WE GET 0=4-(X-1)^2
(X-1)^2=4
X-1=2....OR...X-1=-2
X=3.....OR......-1...
YOU CAN SEE IT FROM GRAPH GIVEN BELOW.

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