# Determine the real numbers a, c if the points (1;2) and (0;3) are on the graph of f(x)=ax^2+x+c.

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### 3 Answers

We are given that the points (1,2) and (0, 3) are on the graph of f(x)=ax^2+x+c.

We can write the following equations:

3 = a*0 + 0 + c = c

and 2 = a +1 +c = a + c + 1

Now from the first equation we have c = 3.

Substitute this in 2 = a +1 +c

=> 2 = a + 1 + 3

=> a = -2

**Therefore the real numbers a and c are a = -2 and c = 3.**

To determine the real numbers a, c if the points (1;2) and (0;3) are on the graph of f(x)=ax^2+x+c.

(1,2) is on ax^2+x+c.

=> f(1) = a*1^2+1+c = 2.

Or a+1+c = 2

a+c = 2-1= 1.

a+c = 1......(1)

(0,3) is on ax^2+x+c.

=> f(0) = a*0^2+0+c = 3.

=> c = 3....(2)

We put c = 3 in (1): a+c = 1. Or a+3 = 1. So a = 1-3 = -2.

**Therefore a = -2, c = 3.**

So ax^2+x+c = -2x^2+x+3.

If the graph of the function f(x) is passing through the given points, then the coordinates of these points have to verify the expression of the function.

The point A(1 , 2) is on the graph if and only if:

f(1) = 2

We'll substitute x by 1:

f(1) = a + 1 + c

a + 1 + c = 2

The point B(0 , 3) is on the graph if and only if:

f(0) = 3

We'll substitute x by 0:

f(0) = c

**c = 3**

We'll substitute c = 3 in:

a + 1 + c = 2

a + 1 + 3 = 2

We'll subtract 4 both sides:

a = 2 - 4

**a = -2**

**The expression of f(x) is: ****f(x) = -2x^2 + x + 3**