- The standard form for a quadratic function is y = ax2 + bx + c where a,b and c are constants and a is not equal to zero.
- The vertex form of a quadratic function is y = a (x - h )2 + k where a, h and k are all constants and a is not equal to zero.
- Case 1: y = a (x - 3 )2 + 2 -- Set (x - 3) = 0 x = 3 and y = 2 Vertex at (3, 2)
- Case 2: y = a (x + 3 )2 + 2 -- Set (x + 3)= 0 x = -3 and y = 2 Vertex at (-3, 2)
- Case 3: y = a (x - 3 )2 - 2 -- Set (x - 3) = 0 x = 3 and y = -2 Vertex at (3, -2)
- Case 4: y = a (x + 3 )2 - 2 -- Set (x + 3) = 0 x = -3 and y = -2 Vertex at (-3, -2)
Case I - You are given the coordinates of the vertex and one other point.
The figure shows that the vertex is at (5, 1) and one other point is known, (1, 4). The vertex form of the quadratic function is f(x) = a(x - h)2 + k. Given that the vertex is at (5, 1) the function can be written as f(x) = a(x -5)2 + 1. To completely define the function we need only to determine the value of a. Using the point (1, 4) we can write 4=a (1 - 5)2 + 1, 4 = a * 16 + 1, a = (4 - 1) / 16 = 3/16. Therefore the function can be written as: f(x) = 3/16(x - 5)2 + 1.
Case II - You are given the coordinates of three points.
In this case the first step is to use the y axis intercept in the standard form of the equation (not the vertex form, since we don’t know where the vertex is located). This will give us 6 = a*02+ b * 0 + c. The solution to this equation is c = 6, so we now know c. Next let’s use the point (8, 6). 6 = 82a + 8b + 6. After simplification this gives 64a + 8b = 0. The other point given is (2, 4.5). Substituting it into f(x) = ax2 + bx + c we get the equation 4.5 = a22 + b 2 + 6. Simplifying the equation we get 4a + 2b = -1.5.
We now have two equations in the variables a and b. These are linear equations and we can solve them for the values of a and b using either substitution or addition/subtraction methods.
Our final relationship is f(x) = 1/8 x2 -(1/2) x + 6
The graph of a quadratic function is called a parabola. When we have the equation in vertex form y = a (x - h )2 + k the vertex of the parabola (high point or low point) we can read the coordinates of the vertex from the equation. It has the coordinates (h, k). Note: To get the x coordinate, with the proper sign, take the terms in the bracket with x {(x - h) or (x + h)} and set them equal to zero. Solve for x. This is the x coordinate of the parabola’s vertex. The y coordinate of the vertex is the value of k (along with its sign). The figures below show the vertex for a parabola that opens up and for one that opens down. If the constant a is positive the parabola will open up. If the constant a is negative the parabola will open down.
- To get familiar with this we will look at four different cases,
When you can easily get your parabola into vertex form it is relatively simple to find the vertex and determine if the parabola opens up or down. To convert from standard form to vertex form you can complete the square (remember that?). The next section briefly discusses completing the square.
Completing the square
Suppose we have a quadratic function in standard form and want to covert it to vertex form. Consider the function m(x) = x2 + 6x . We want to make it into a perfect square. If we took (x + a)2 and squared it we would get: x2 + 2ax + a2. The middle term is 2ax. Our middle term is 6x. If we want a perfect square 2ax must equal 6x. Solving for a we get a = 6/2 = 3. The last term in a perfect square is a2 , which is this case is 32 or 9. To make our equation into a perfect square we must add a2 (9), but you can’t just add an amount to a function as this changes the function. So we must add a2 and subtract it, so the net change is 0. Thus we get m(x) = x2 + 6x + 9 - 9 . Rewriting this into vertex form we get m(x) = (x + 3)2 - 9. From the vertex form of the function we can tell that the vertex of our parabola is at (-3, -9) and the parabola opens up.
What if you cannot easily complete the square?
If you have a quadratic function in the form m(x) = ax2 + bx + c and the coefficients are not nice to work with, making completing the square a difficult task you can use the relationship that we will derive next. Remember the quadratic formula? If you don’t it looks like this:
The quadratic formula is used to find the roots or zeroes of a quadratic function. The roots (also called zeroes) represent the places where the function crosses the x axis. The vertex is mid-way between the two x axis crossings.
We are going to use the quadratic formula to find the x values at a and b and the point mid-way between them (x of vertex). We will use the quadratic formula to do this:
So, if you have a quadratic function in standard form, and it's really hard to complete the square you can find the x of the vertex from the equation x = -b/2a. Where a is the coefficient of the x squared term and b is the coefficient of the x term. For example assume you have the function u(x) = 1.5x2 + 4.5 x - 8. a = 1.5, b = 4.5 and the x coordinate of the vertex is x = -4.5/(2*1.5) = -4.5/3 = -1.5. To find the y coordinate of the vertex you substitute your x vertex value into the function and calculate as follows: y = 1.5(-1.5)2 + 4.5 (-1.5) -8 = 3.375 - 6.75 - 8 = -11.375 So, the x,y coordinates of the vertex are (-1.5, -11.375)
What if I have to determine the function when all I have is the graph?
We will look at two cases. Case I will involve the situation where the vertex and one other point are given on the graph. Case II will involve the problem when you are given three points (to make life easy one of them is the y axis intercept).