Blog Post 4

Hello, my name is Camila Haldemann. This class we will be discussing quadratic functions, I know they can be a bit tricky compared to linear functions, but once you understand them, they will be easy.  A quadratic function is a second-degree polynomial function, and it has a u shape (parabola). Quadratic functions can be written in three different ways: in standard form, vertex form, and factored form.

The standard form is F(X)= ax²+bx+c, where a,b,c are constants. Just by looking at this formula one can learn many things. The first is that if “a” the leading coefficient is negative, then the graph of F(X) would be concave down, and therefore due to the u shape, it would have a maximum value because the graph would have a sad face shape. On the other hand, if “a” the leading coefficient is positive, the graph of F(X) would be concave up, and therefore it would have a minimum value due to the happy face shape the graph would have. Adding to these there are other details that can be known by looking at the equation, and these are if the graph would be horizontally compressed or stretched, this means how wide the graph would look. If a > 1, then the graph would be compressed meaning that it would be close to the y-axis, however if 0<a<1 then the graph would be stretched horizontally, meaning that it would be closer to the x-axis.

Continuing the vertex form is F(X)= a (x-h)²+k, where (h,k) is the vertex. In a parabola or quadratic function, the vertex is the starting point. This means that the coordinate (h,k) is the starting point of the function. Are you following me?

Next, is the factored form which is F(X)= a(x-r)(x-s), where “r” and “s” are the x-intercepts or zeroes of F(X), in other words, this is saying that “r” and “s” would be the coordinates of the points that are crossing through the x-axis.

An example of an exercise of quadratic functions would be the next.

Find the formula of a quadratic function that has the x-intercepts (0,1) (0,3) and the y-intercepts (6,0). As you are given the x-intercepts, you can input this in the factored form function. It would look like this: F(X)= a (x-1)(x-3). Now, as you also have the y-intercept you can substitute it in the equation, 6=a(0-1)(0-3), and later solve for a. 6=a(-1)(-3), 6=3a, a=2. Finally, you have all the pieces for a function, it would look the following way in factored form: F(X)= 2(x-1)(x-3). However, you could also change it into standard form, by doing distributive property. In standard form it would look like this: F(X)= 2X²-8X+6.

Another example is the following.

Find an equation in vertex form of a parabola that passes through (3,-6) and has a vertex of (1,2).

You know the equation in vertex form is F(X)= a (x-h)²+k. Therefore, you would just substitute the vertex coordinate into the equation. F(X)= a(x-1)²+2. Next, you would substitute the other coordinate (3,-6) into the remaining x and y in the equation. -6=a(3-1)²+2. Finally you would solve for a. -6=a(4)+2, -8=4a, a=2. In the end, the final equation in vertex form would be, F(X)=-2(x-1)²+2. Based on this equation we would know that as the leading coefficient “a” is negative, then it would be concave down having a maximum value.