Formulate the quadratic function that contains the points (-2,1),(0,1)and (1,4)

  y = x^2 +2x +1

Step-by-step explanation:

Let's see how this works.

Two of the given points have the same y-value, so the axis of symmetry is located halfway between. If we translate the desired curve down 1 unit, those symmetrical values become zeros, and our function will look like ...

  f(x) = ax(x +2)

We want this to go through the third point, also translated down by 1 unit, so we have ...

  3 = a(1)(1+2) = 3a

That is, a=1, so f(x) = x(x+2) and translating that up 1 unit gives our desired function:

  y = x(x+2) +1

  y = x^2 +2x +1

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Using the quadratic regression function of a graphing calculator gives the same result.

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Comment on the solution approach

For arbitrary points, one would generally substitute them into the formula for  the function you want:

  y = ax^2 +bx +c

to get three equations in the parameters a, b, and c. These equations would be solved to find the desired coefficients. As noted elsewhere, one of the given points for this problem is the y-intercept (0, 1), so right away we know c = 1. ("c" is the y-intercept.)

That leaves two remaining equations in two unknowns (using the other two points).

Using the point (-2, 1), the equation is ...

  1 = a(-2)^2 +b(-2) +1   ⇒   4a - 2b = 0   ⇒   2a -b = 0

Using the point (1, 4), the equation is ...

  4 = a(1^2) +b(1) +1   ⇒   a +b = 3

Adding these gives 3a = 3, or a=1. Then substitution gives b=2, so the final equation is ...

  y = x^2 +2x +1