hELP WITH MATH??

Question

i have to find the vertx, x and y intercepts and axis and then draw the graph of y=-x^2+4x

PLEASE HELPP

Answer 1

use your TI-83 graphing calculator!

Answer 2

Pick values for x...like 0, 1, 2, -1and solve for y. Use those values to plot the graph. That will show what the x-intercept is (where the graph crosses the x (or horizontal axis) and where the y-intercept is (where the graph crosses the vertical axis).

For example:
If x=0, then y=-(0)^2 +4*(0)
then solving for y gives you y=0
Put your first point on your graph at x=0, y=0 (also called the origin).

Keep going until you can connect to lines to see what the final graph looks like(it will be a parabola). The vertex is at the base of the parabola (where the two sides come together). You can also pick this point off of your graph.

Answer 3

First, any x^2 graph is a parabola

x intercepts:
set -x^2 + 4x = 0
x(-x + 4) = 0
so, x intercept at x = 0 and x = 4

y intercepts: at (0, 0)

Calculus Warning ....
Derivative , y = -2x + 4, graphs the rate of change
-2x + 4 = 0, set to 0 to find when the rate of change is 0, or the maximum of the parabola

-2x = -4
x = 2

Now, evaluate the graph at x = 2 and you will get y = 4, so the low point of the graph is at (2, 4)

Also, remember because the graph is negative, it is pointing downward like an upside down smiley face.

Answer 4

Need to take the derivative and set to zero to find max/min point or the point where the slope is zero. At this point the curve changes direction. dy/dx = 2x + 4 . At dy/dx (0) = -2 . The point (-2,-4) is the bottom of the curve. Set x = 0 to find y intercept which is 0. There are two x intercepts 0 and -4.