Question

Write a quadratic function in standard form whose graph passes through the given points1. (-1,5), (0,3), (3,9)

Answer 1

The Standard Form of a Quadratic Function:

$\text{ y = ax}^2\text{ + }bx\text{ + c}$

Using the given points (-1,5), (0,3), and (3,9)​, let's substitute each point to the equation.

At (-1,5):

$\text{ y = ax}^2\text{ + bx + c }\rightarrow5=a(-1)^2\text{ + b(-1) + c}$$\text{ 5 = a - b + c}$

At (0,3):

$\text{ y = ax}^2\text{ + bx + c }\rightarrow3=a(0)^2\text{ + b(0) + c}$$\text{ 3 = c}$

At (3,9):

$\text{ y = ax}^2\text{ + bx + c }\rightarrow9=a(3)^2\text{ + b(3) + c}$$\text{ 9 = 9a + 3b + c}$

We now get these equations:

5 = a - b + c ; 3 =c; 9 = 9a + 3b + c

Let's determine the value of a, b and c. We get,

Substituting 3 = c to 5 = a - b + c,

$\text{ 5 = a - b + c }\rightarrow\text{ 5 = a - b + 3 }\rightarrow\text{ a - b = 2}$$\text{ b = a - 2}$

Let's substitute 3 = c and b = a - 2 to 9 = 9a + 3b + c,

$\text{ 9 = 9a + 3b + c }\rightarrow\text{ 9 = 9a + 3(a-2) + 3}$$\text{ 9 = 9a + 3a - 6 + 3 }\rightarrow\text{ 12a = 9 + 6 - 3 }\rightarrow\text{ 12a = 12}$$\text{ a = }\frac{12}{12}\text{ = 1}$

Since a = 1, let's solve for the value of b which is b = a - 2.

$\text{ b = a - 2 }\rightarrow\text{ b = 1 - 2}$$\text{ b = -1}$

Since we've identified that a = 1, b = -1 and c = 3, let's substitute the values to the standard form of a quadratic function to be able to make the equation.

$\text{ y = ax}^2\text{ + bx + c }\rightarrow y=(1)x^2\text{ + (-1)x + (3)}$$\text{ y = x}^2\text{ - x + 3}$

Therefore, the quadratic function in a standard form whose graph passes through the given points (-1,5), (0,3), (3,9)​ is y = x^2 - x + 3.