SOLUTION: Could someone please show me how to do this? Find the x-intercepts. {{{y=x^2+4x}}}

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Question 102747: Could someone please show me how to do this? Find the x-intercepts. y=x%5E2%2B4x
Found 2 solutions by bucky, MathLover1:
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
An x-intercept is a point on the x-axis where the graph of the function crosses the x-axis.
But any (x,y) point on the x-axis has zero as its y-value. So to find the intercept
points on the x-axis, set y equal to zero and solve for the corresponding values of x. So
let's do that ... set y = 0 in the given function of y+=+x%5E2+%2B+4x. When you do that
you get:
.
0+=+x%5E2+%2B+4x
.
Let's transpose this ... switch sides around ... to get it in the little more standard form
of:
.
x%5E2+%2B+4x+=+0
.
Since the two terms on the left side both contain x, we can factor an x from each of the
terms to make the equation become:
.
x%2A%28x+%2B+4%29+=+0
.
This equation will be true if either one of the factors equals zero because multiplying
the left side by zero will make it equal the right side which is zero.
.
So the equation will be true if either x = 0 or x + 4 = 0. In the second factor this means
that x = -4. Since we already know that y is zero for these two values, the x-intercept
points are (0, 0) and (-4, 0).
.
That's all there is to it.
.
Hope this helps you to understand that to find x-intercepts you set y equal to zero and solve
for the corresponding values of x.
.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

The x-intercepts are where the graph crosses the x-axis, and the y-intercepts are where the graph crosses the y-axis.
y+=+x%5E2+%2B4x
Find x for what y = o
x%5E2+%2B+4x+=0
x%28x+%2B+4%29+=+0 this product is equal to 0 if x+or %28x%2B4%29 is equal to 0
Solutions:
x1+=+0
x2+=+-4
Here is the graph:


Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B0+=+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=%284%29%5E2-4%2A1%2A0=16.

Discriminant d=16 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+16+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+16+%29%29%2F2%5C1+=+0
x%5B2%5D+=+%28-%284%29-sqrt%28+16+%29%29%2F2%5C1+=+-4

Quadratic expression 1x%5E2%2B4x%2B0 can be factored:
1x%5E2%2B4x%2B0+=+1%28x-0%29%2A%28x--4%29
Again, the answer is: 0, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B0+%29