SOLUTION: the problem is 3x^+x-5 I need to find the x intercept and the vertex . I am not sure how to set up the equation or what numbers go where. HELP PLEASE

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Question 162673: the problem is 3x^+x-5 I need to find the x intercept and the vertex . I am not sure how to set up the equation or what numbers go where. HELP PLEASE
Answer by vleith(2983)   (Show Source): You can put this solution on YOUR website!
See this for a fairly detailed look at this issue --> http://www.analyzemath.com/quadratics/vertex_problems.html
I assume you meant
Any point that is an x axis intercept must have a y coordinate value of 0. So to solve for those points. solve
You can use quadratic equation for that.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=61 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 1.13504161265111, -1.46837494598444. Here's your graph:


The URl above gives you a formula for finding the vertex. Here are two others ways
1)if you know deriviatives, take the first derivative and set that to 0. Then solve for x. Use that x vlaue to find the corresponding y in the original equation.
2) find the x value that is the average of the 2 x intercepts. That point is on the line that 'splits' the parabola (the line about which the parabola has symmetry). Find that x value and plug it into the original equation to find the y value.
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