help me pleaseee

Question

The points (0, 3), (1, 4), (2, 7), and (3, 12) are on the graph of a function.

Write an explicit function that can be used to represent the function.

Answer 1

Hi Christina B.

You can check the points as given against the equations for the slope of a line, since you have four points.

If the slopes are not the same the you don't have a line

Slope m = (y₂ - y₁)/(x₂ - x₁)

(4-3)/(1-0) = 4

(7-4)/(2-1) = 3

(12-7)/(3-2) = 5

Next you can use the Standard Form of a Parabola versus three of the points you have see if you have a parabola which would a 2nd degree equation ax² + bx + c = f(x) = y

for (0,3)

3 = a(0)² + b(0) + c

3 = c

So we have a value for c

for (1,4)

4 = a(1)² + b(1) + c

4 = a + b + c

Plug in the value we have for c

4 = a + b + 3

1 = a + b

For (2,7)

7 = a(2)² + b(2) + c

7 = 4a + 2b + c

Plug in the value for c

7 = 4a + 2b + 3

4 = 4a + 2b

We have two equations in two unknowns a and b they are

a + b = 1

4a + 2b = 4

If we multiply, a + b = 1 by negative 2 we can eliminate b

-2a - 2b = -2

4a + 2b = 4

We get

2a = 2

a = 1

Now we have values for a and c and we can plug these back into any of the original equations to find b

For ease I will use this one

4 = a + b + c

4 = 1 + b + 3

4 = 4 + b

0 = b

We have values for all the variables

a = 1, b = 0 and c = 3

So

ax² + bx + c = x² + 0 + 3 = x² + 3

f(x) = x² + 3

You can graph this at Desmos.com

You confirm that (0,3), (1,4), (2,7) and (3,12) are all points on your parabola

I hope this was useful

Answer 2

The first thing that you would want to do when you approach this problem is to graph the points you are given to see what your graph looks like. This will tell us what kind of function we are working with. For instance, if it were a straight line, we would know that the function is linear.

In this case, the shape of the graph is exponential. So we know that our function has to include x to the power of some number. Now, we can look at one of our more valuable points: the y-intercept. This is given by our prompt: (0,3). This means that our function has to include the term "+ 3." So, given this, we can solve for what our exponent should be.

I will use the letter n to represent our unknown exponent in the following formula template: f(x) = xⁿ + 3.

From here, we can plug in our values. We shouldn't plug in our y-intercept point because 0ⁿ is always 0, so we don't learn anything from it. Similarly, our point (1,4) won't tell us anything because 1ⁿ is always 1. We can, however, use our third point of (2,7). When we plug it in, we have 7 = 2ⁿ + 3 which simplifies to 2ⁿ = 4. We know that 2² is equal to 4, so we can find that our n value is equal to 2. This can be confirmed with the final point which, when plugged in, simplifies to 3ⁿ = 9, showing that our n value must be 2.

The explicit function that can represent the graph of the function is then f(x) = x² + 3. Let me know if you have any questions on this process!