SOLUTION: I'm having trouble with this text book question can you show me how to solve it graph parabola with vertex and intercepts f(X)= xsquared+6X+5 (sorry I dont know how to make

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I'm having trouble with this text book question can you show me how to solve it graph parabola with vertex and intercepts f(X)= xsquared+6X+5 (sorry I dont know how to make      Log On


   



Question 22916: I'm having trouble with this text book question can you show me how to solve it
graph parabola with vertex and intercepts
f(X)= xsquared+6X+5 (sorry I dont know how to make a small 2 after the first x)

Found 2 solutions by NaiXLG, Earlsdon:
Answer by NaiXLG(18) About Me  (Show Source):
You can put this solution on YOUR website!
So, you want to solve the equation;

f%28x%29+=+x%5E2+%2B+6x+%2B+5
and then graph it....
Well first set it up
0+=+x%5E2+%2B+6x+%2B+5
And then factor....you have to ask yourself...what two factors of 5 add up to six? Well, there are ONLY two factors of 5, that is 1 and 5 (because 1 * 5 = 5). So, go ahead and factor;
0+=+%28x%2B1%29%28x%2B5%29
***If you foil that, you'll get back to the original equation
Now use something called the zero product property, which means set the equations inside of those parentheses to zero, and solve;
x%2B1+=+0 & x%2B5+=+0
x+=+-1 & x+=+-5
Those are the solutions...
Now, to find the vertex, use the formula;
x+=+%28-b%2F2a%29
So, we are going to plug in the values of the a, b and c from the orignal equation. Remember +ax%5E2+%2B+bx+%2B+c+
+x+=+%28-6%2F2%281%29%29+
+x+=+-3+

Now, finally, here is the graph...

graph%28300%2C200%2C-7%2C3%2C-5%2C10%2Cx%5E2%2B6x%2B5%29

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Ok, you can show x squared like this...x^2 by using the caret (^). You get the caret by Shift-6 key. A little advice...use the same type letter (all upper case or all lower case) for the variable in your problems, rather than swithichjng from X to x and back again.
Solve: You can also write your equation: y+=+x%5E2+%2B+6x+%2B+5
y+=+x%5E2+%2B+6x+%2B+5 You can solve this by factoring if you set y = 0
%28x+%2B+1%29%28x+%2B+5%29+=+0 Apply the zero product principle.
%28x+%2B+1%29+=+0 and/or %28x+%2B+5%29+=+0
If x+%2B+1+=+0 then, x+=+-1
If x+%2B+5+=+0 then, x+=+-5
The x-intercepts (also called zeros) are:
x+=+-1
x+=+-5
The x-coordinate of the vertex is found by:
x+=+-b%2F2a The a and b come from the standard form of the quadratic equation: ax%5E2+%2B+bx+%2B+c In your equation, a = 1 and b = 6
x+=+-6%2F2
x+=+-3 This the x-coordinate of the vertex. Find the y-coordinate by substituting this value of x into your equation and solving for y.
y+=+x%5E2+%2B+6x+%2B+5 x = -3
y+=+%28-3%29%5E2+%2B+6%28-3%29+%2B+5
y+=+9+-+18+%2B+5
y+=+-4
The vertex is at (-3, -4)
Here's the graph.
graph%28300%2C200%2C-10%2C3%2C-5%2C5%2Cx%5E2%2B6x%2B5%29