- Submitted By: olive6991
- Date Submitted: 11/03/2013 2:03 PM
- Category: Science
- Words: 357
- Page: 2
- Views: 65

Quadratic Essay

f(x)= 2x^2 + 8x + 5

Vertex: In order to find the vertex of the quadratic equation, begin by using the proper formula - b / 2(a) to find it. In the equation there is an (a) (b) (c) which is needed for the vertex equation. First look at the equation and determine what are the values of the three variable. In this case (a)= 2, (b)= 8, (c)= 5. Now plug them in properly into the vertex equation. * Notice in the vertex equation there is a negative sign!!! DO NOT FORGET! When plugged into the equation, it should state -8 / 2(2). Now use the proper steps and multiply the denominator by 2 as it implies resulting the denominator equalling 4. Now the vertex equation should be (-8 / 4) plus reducing would make it (-2). The (x) value of the vertex (-2, 0). To find the (y) value plug in the x-vertex back into the original equation f (x)= 2x^2+8x+5, resulting to be f(x)= 2(-2)^2+8(-2)+5. Use the order of operation (PEMDAS) to solve. * Do not forget that the exponent (^2) gets multiplied into (-2) not the product of 2(-2). After using those steps the f(x) or (y) value should equal -3, making the vertex (-2, -3).

x-intercept: The second step is finding the x-intercept (zeros, roots). Begin by using the quadratic formula to find the x-intercept. As stated in the process of finding the vertex, use the same variables and values (a)= 2, (b)= 8, (c)= 5. Plug in the values into the equation

Using the same order of operation it should be

then reduce,

Now to find the point make

add and subtract and

equalling (.75 , -3.75).

Graphing: Having the vertex point and the x-intercept points from the quadratic equation plug in the results into a graphing calculator to check if the answers are correct. Check if your vertex point (-2, -3) and x-intercepts (.75, -3.25) match to your graph.

* points of the x-intercept fall on the x-axis creating a parabola graph.