Question 297565: Graph the quadratic function
f(x) = -x2 + 1
Describe the correct graph and why...Which way does the parabola go up or down, where does the graph cross the x-axis and y-axis....
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Graph the quadratic function
f(x) = -x^2 + 1
Describe the correct graph and why...Which way does the parabola go up or down, where does the graph cross the x-axis and y-axis....
The graph points up and opens down.
Standard form of quadratic equation is ax^2 + bx + c = 0
Set f(x) = to 0 and you have this equation in standard form.
You get:
-x^2 + 1 = 0
In this equation:
a = -1
b = 0
c = 1
Maximum point is at x = -b/2a which becomes 0
When x = 0, y = 1, so the maximum point is (x,y) = (0,1).
To find the points where this graph crosses the x-axis, you have to solve the equation -x^2 + 1 = 0
With this equation, you subtract 1 from both sides of the equation to get:
-x^2 = -1
Multiply both sides of this equation by -1 to get:
x^2 = 1
Take the square root of both sides of this equation to get:
x = +/- 1
Those should be the x-axis crossing points.
You could also have factored the equation of -x^2 + 1 = 0 to get:
(-x+1) * (x+1) = 0
When either of these factors = 0, the equation is grue, so you set each of the factors equal to 0 and solve.
You get:
x = -1 and x = 1.
You graph this equation by plotting some values of x and getting corresponding values of y.
You start with x = -1, x = 0, x = 1
That should be enough to draw a rough graph, but you might want to fill in some additional points to fit the curve better.
Your graph should look like this:
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