Write a quadratic in standard form using standard form. NO CALCULATOR (-1,2) , (-2,7) , (0,7)

Question

Write a quadratic in standard form using standard form. NO CALCULATOR

 

(-1,2) , (-2,7) , (0,7)

 

Answer 1

We can also create a system of equations using these points. You know that the standard form of a quadratic function is

 

y = ax² + bx + c

 

 

Since the point (0, 7) is the y-intercept, we can say

 

c = 7                     eq1

 

Then we use the other second point (-2, 7):

 

(-2)²a + (-2)b + 7 = 7

 

4a - 2b = 0              eq2

 

The last point (-1, 2)

 

(-1)²a + (-1)b + 7 = 2

 

a - b + 7 = 2

 

a - b = -5                 eq3

 

 

The bolded equations become your system of equations.  Substitute eq3 into eq2.

 

4(-5 + b) - 2b = 0

 

Solving for b results in

 

-20 + 4b - 2b = 0

 

 2b = 20

 

  b = 10

 

 

Substitute this value of b into eq2 to solve for a.

 

a = -5 + b

 

a = -5 + 10

 

a = 5

 

 

So your quadratic function is

 

y = 5x² + 10x + 7

Answer 2

First off, notice that two of the points have the same y value  (-2,7) and (0,7).  Since a parabola is symmetric, the axis of symmetry must lie exactly midway between them, so x = (-2 + 0)/2 = -2/2 = -1

 

By a stroke of good luck the third point (-1,2) lies on the axis of symmetry and must therefore be the vertex.

 

the equation must be

 

y = a[x - (-1)]² + (2) = a(x + 1)² + 2

 

since (0,7) is a given value we can solve for a

 

(7) = a [(0) + 1]² + 2

5 = a(1)

 

so so the equation is

 

y = 5(x + 1)² + 2

 

Check with the other known value (-2,7)

7 = 5[(-2) + 1]² + 2

5 = 5 (-1)² = 5(1) = 5