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
Explanation:
Since the
#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 = 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#
Explanation:
Alternatively, consider the general problem of finding the equation of a parabola with vertical axis passing through points
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
Since
Then
So
Then
Hence
#=(y_3-y_1)/((x_3-x_1)(x_3-x_2)) - (y_2-y_1)/((x_2-x_1)(x_3-x_2))#
Turning to our example:
Plugging these values into the formulae for
#{ (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#
Explanation:
We want to find a quadratic function
Since
So
Then from
#-2 = a(1-5/2)^2 + b = 9/4 a + b#
and from
#-4 = a(2-5/2)^2 + b = 1/4 a + b#
Subtracting this second equation from the first we find:
#2a = 2#
So
and
So:
#f(x) = (x-5/2)^2-17/4 =x^2-5x+25/4-17/4 = x^2-5x+2#