We want to determine whether there is a quadratic function that fits the given set of values.

To use the given points, we need to substitute their coordinates into the standard form of a quadratic function.

y = a x^{2} + b x + c, a \neq = 0

Doing so will create a system of equations that we can solve for the values of

a,

b,

and

c .

Let's do it!

We already found that

c = 0 .

This allows us to start writing our equation.

y = a x^{2} + b x + 0 ⇕ y = a x^{2} + b x

Now we can write a system using the two remaining equations.

{4 a - 2 b + c = 16 a + b + c = 4 ⇓ {4 a - 2 b + 0 = 16 a + b + 0 = 4 (I) (II)

Let's solve this system using the Elimination Method. First, we will simplify the equations, then we will divide Equation (I) by

2,

and finally we will add Equation (II) to Equation (I).

{4 a - 2 b + 0 = 16 a + b + 0 = 4 (I) (II)

▼

(I), (II):

Simplify

$(I), (II):$ Identity Property of Addition

{4 a - 2 b = 16 a + b = 4

$(I):$ Factor out $2$

{2 (2 a - b) = 16 a + b = 4

$(I):$ $LHS / 2 = RHS / 2$

{2 a - b = 8 a + b = 4

$(I):$ Add $(II)$

{2 a - b + a + b = 8 + 4 a + b = 4

▼

(I):

Solve for

a

$(I):$ Add terms

{3 a = 12 a + b = 4

$(I):$ $LHS / 3 = RHS / 3$

{a = 4 a + b = 4

We found that

a = 4 .

y = 4 x^{2} + b x

Next, we will substitute

4

for

a

in Equation (II).

{a = 4 a + b = 4 (I) (II)

$(II):$ $a = 4$

{a = 4 4 + b = 4

$(II):$ $LHS - 4 = RHS - 4$

{a = 4 b = 0

Now that we have the three values, we can write the full equation of the quadratic function.

y = 4 x^{2} + 0 x ⇕ y = 4 x^{2}

To help visualize this situation, we can plot the given points and sketch the curve.

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