Question 297565
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:


{{{graph(600,600,0-5,5,-5,5,-x^2+1)}}}