Quadratic Formula Lab Report

780 Words 4 Pages
In order to derive the Quadratic Formula by completing the square, it was important to verify that the quadratic equation was in standard form which is ax? + bx + c = 0. By using this correct form, I prevented myself from using the wrong values for a,b, or c which could result in an incorrect solution. I began by rewriting the equation by subtracting c to the other side which isolated ax? + bx. I divided the equation by a because it was multiplied on the squared term. In addition, I took the half of the x-term and squared it, adding that squared term to both sides: + =+ To make the work more organized, I simplified the right-hand side by simply converting to a common denominator: x? + + = +. After this, I converted the left-side to square form plus a bit of simplifying on the right:
= From this point, I began to square root both sides and remembered to put the sign on the right: x.
Once I did that, I solved for x and simplified as necessary: x=
By evaluating the answer, it finally resulted in the Quadratic Formula:

Problem 2: 2x?-x+2=0 Rewriting in vertex form: Before I rewrote the quadratic equation of 2x?-x+2=0 in vertex form, I had to understand that the vertex form of a quadratic function is f(x)=a(x-h)?+k where
…show more content…
A parabola will always have a domain of all x-values so I immediately knew that the domain of this quadratic equation is all real numbers, or (?∞,∞). To find the range, I evaluated the (h,k) which in this equation is (1/4, 15/8). I can see that this parabola has a minimal value of 15/8 and go up to positive infinity. The range is y is greater than or equal to 15/8, or y. The maximum/minimum value can be found by the output of the quadratic function at the vertex. By evaluating the vertex in y=2(x-1/4)?+15/8, I noticed that -1/4 was a negative number which meant that it had a minimum value occurring at

Related Documents