SOLUTION: f(x)=25xsquare2+20x+4 find the y intercepts and x intercepts

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Question 491888: f(x)=25xsquare2+20x+4 find the y intercepts and x intercepts
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Intercepts can be found by plugging in zero for the other variable since anywhere on y-axis, x=0 and anywhere on x-axis, y=0
Y-intercept: set x=0
--> y = 0+0+4 = 4
Y-intercept is the point (0,4), this is where the graph crosses the y-axis
X-intercept: set y=0
--> 25x^2 +20x +4 = 0
I will use the quadratic solver to show how to find x:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 25x%5E2%2B20x%2B4+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2820%29%5E2-4%2A25%2A4=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%2820%29%29%2F2%5C25.
Expression can be factored: 25x%5E2%2B20x%2B4+=+25%28x--0.4%29%2A%28x--0.4%29

Again, the answer is: -0.4, -0.4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+25%2Ax%5E2%2B20%2Ax%2B4+%29