What is the value of the x-coordinate of the vertex of the function shown below? f(x) = (x + 2)(x - 6) I converted this over to standard form and I almost have it in vertex for, which is what I need it in to complete this, I'll post my work below, I just need help with one step...
yeah thats one way to work the x coordinate of vertex, but there is a shortcut if you use symmetry : `x coordinate of vertex = average of x intercepts`
f(x) = x^2 - 6x + 2x - 12 f(x) = x^2 - 4x - 12 y = (x - 6)(x + 2) Because I pulled these from the equation, the equation is correct... Would the y just be equal to this? How would yhu do it with the symmetry?
f(x) = x^2 - 4x - 12 looks good, convert it into vertex form
do you know how to complete the square ?
No, I don't. I haven't heard of completing a square...
have you heard of -b/2a ?
OH! Yea. I know how to do it. c:
Well you have it factorized, so you know the x intercepts, 6 and -2 by using the null-factor law. Parabolas are symmetrical, which means one side is the mirror-image of the other and that the x co-ordinate of the vertex is exactly half way between intercepts. To find the midpoint, add the intercepts and half the sum, which is what @ganeshie8 said. 6-2=4, 4/2=2 Now to find the y co-ordinate, just plug it back into the original function.
good work it using -b/2a and see if it matches with the method described by ilr1, please medal ilr1 for nice explanation :) i gave u medal already and don't like to remove...
Will do. c: So, with x being 2, I plug it in... So then it would be f(2) = (2 + 2)(2 - 6) Then... I foil it? ...... Right?
>.< I confused myself.
f(x) = x^2 - 4x - 12 a = 1 b = -4 x coordinate of vertex = -b/2a = (--4 )/ 2(1) = 4/2 = 2
which is same as the value worked out by ilr1 using symmetry
Yea, I got that far. . . Wait, If I was to plug it into the original equation, would that be checking it.... So 2 is my final answer?
2 is the x coordinate of vertex and yes thats the final answer
if you want y coordinate, plugin x=2 in the equation
Oh! I see now. Thank yhu both so much for the help ^.^ I appreciate it.
np :)
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