Given three points (1,-2), (2,-4) and (3,-4), how do you write a quadratic function in standard form with the points?

3 Answers
Apr 7, 2016

#f(x) = x^2 -5x + 2#

Explanation:

Since the #x# coordinates of the points are #1#, #2# and #3#, this problem is equivalent to finding a quadratic formula for the terms of the sequence:

#color(blue)(-2), -4, -4#

Write down the sequence of differences of this sequence:

#color(blue)(-2), 0#

Write down the sequence of differences of this sequence:

#color(blue)(2)#

Then we can construct a formula for #a_n# (#n = 1, 2, 3#) using the initial term of each of these sequences as coefficients:

#a_n = color(blue)(-2)/(0!) + color(blue)(-2)/(1!)(n - 1) + color(blue)(2)/(2!)(n-1)(n-2)#

#=-2-2n+2+n^2-3n+2#

#=n^2-5n+2#

This corresponds to the standard quadratic function:

#f(x) = x^2-5x+2#

Apr 7, 2016

#f(x) = x^2-5x+2#

Explanation:

Alternatively, consider the general problem of finding the equation of a parabola with vertical axis passing through points #(x_1, y_1)#, #(x_2, y_2)# and #(x_3, y_3)#.

Consider a definition of the form:

#f(x) = a + b (x-x_1) + c (x-x_1)(x-x_2)#

where we want to determine #a#, #b# and #c#.

Since #f(x_1) = y_1# we must have #a = y_1#

Then #y_2 = f(x_2) = y_1 + b(x_2 - x_1)#

So #b = (y_2 - y_1)/(x_2 - x_1)#

Then #y_3 = f(x_3) = y_1 + (y_2 - y_1)/(x_2 - x_1)(x_3-x_1) + c(x_3-x_1)(x_3-x_2)#

Hence #c = (y_3 - y_1- (y_2 - y_1)/(x_2 - x_1)(x_3-x_1))/((x_3-x_1)(x_3-x_2))#

#=(y_3-y_1)/((x_3-x_1)(x_3-x_2)) - (y_2-y_1)/((x_2-x_1)(x_3-x_2))#

#color(white)()#
Turning to our example: #(x_1, y_1) = (1, -2)#, #(x_2, y_2) = (2, -4)# and #(x_3, y_3) = (3, -4)#

Plugging these values into the formulae for #a, b, c# we find:

#{ (a = -2), (b = -2), (c = 1) :}#

So:

#f(x) = -2-2(x-1)+(x-1)(x-2)#

#=-2-2x+2+x^2-3x+2#

#=x^2-5x+2#

Apr 7, 2016

#f(x) = x^2-5x+2#

Explanation:

We want to find a quadratic function #f(x)# passing through the points #(1, -2)#, #(2, -4)# and #(3, -4)#.

Since #f(2) = f(3) = -4# the vertical axis must lie half way between these #x# coordinates, at #x = 5/2#.

So #f(x) = a(x - 5/2)^2 + b# for some #a# and #b#.

Then from #f(1) = -2# we get:

#-2 = a(1-5/2)^2 + b = 9/4 a + b#

and from #f(2) = -4# we get:

#-4 = a(2-5/2)^2 + b = 1/4 a + b#

Subtracting this second equation from the first we find:

#2a = 2#

So #a=1#

and #b = -2 - 9/4a = -2 - 9/4 = -17/4#

So:

#f(x) = (x-5/2)^2-17/4 =x^2-5x+25/4-17/4 = x^2-5x+2#