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Find the x intercept and coordinates of the vertex for the parabola y=x^2+2x-35

User Sourcerer
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The x-intercepts are found when y is set to equal 0. You then factor the polynomial to find these x values. The vertex is found by completing the square on the polynomial. We will first complete the square to find the vertex, then continue with the process and factor it. Completing the square is in fact a method that can be used to factor. We will first set the equation equal to 0 and then move the constant to the other side of the equals sign to get this:
x^2+2x=35. Now take half the linear term, square it, and add that number to both sides. Our linear term is 2. Half of 2 is 1, and 1 squared is 1, so we will add 1 to both sides.
x^2+2x+1=35+1. What we have done in this process so far is created a perfect square binomial on the left. We will simplify to that binomial and at the same time do the addition on the right:
(x+1)^2=36. If we move the 36 over by subtraction we have the coordinates of the vertex. It is (-1, -36). Now, as tedious as it may seem, move the 36 back over by adding to get back to
(x+1)^2=36. In order to solve for x, we will take the square root of both sides. Doing that gives us this:
x+1=+/-6. Subtract the 1 from both sides and you have your 2 x-intercepts.
x=-1+6 which is x = 5, and
x=-1-6 which is x = -7. To sum up, your vertex is (-1, -36) and your x-intercepts are x = 5 and x = 7.
User Nefas
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