Week 9 – Quadratic Functions

Posted on by shannonf2015

Throughout the week, I have learned how to convert a quadratic function in general form to standard form and factored form to be able to find different information about how the graph looks like.

Let’s start with an example:

y=2x^2-20x+32

Right away, we can tell that the y-intercept is 32 because it is in general form.

We can also tell the direction of opening as it is positive, which means it opens up and has a minimum value.

Start by removing the greatest common factor which is 2.

y=2(x^2-10x+16)

Next, factor the three term in the bracket to find the x-intercepts (the roots)

y=2(x-2)(x-8)

x intercepts: 2, 8

Now we need to find the vertex, to do that we need to change the general form to standard form by completing the squares. We cannot change from factored form to standard form.

y=2x^2-20x+32

First take out what is common between the first two terms, leaving the 32. Leave two spaces for a + value and – value between middle term and last term to create zero pairs.

  • y = 2(x^2-10x+___-___) + 32

Take middle term and divide it by 2 and square it.

  • -\frac{10}{2}=5 —> (-5)^2=25
  • y=2(x^2-10x+ 25-25) +32

Multiply the coefficient (2) with the negative number of the zero pair (-25). Then do the calculations with the last term.

  • y=2(x^2-10x+25)+2(-25)+32
  • y=2(x^2-10x+25)-50+32
  • y=2(x^2-10x+25)-18
  • y=2(x-5)^2-18

Standard form: y=a(x-p)^2+q

Vertex: (p,q)

The vertex is (5, -18); the value of p changes sign when written for the vertex.

Now that we know the vertex, we can determine the axis of symmetry, domain and range.

AOS = x=5

Domain: xER

Range: y ≥ -18

This equation is also congruent to y=2x^2.

Another example is when we are given some information about the quadratic function and we must write an equation.

Example: The graph of a quadratic function passes through (1, -7) and the zeros of function are -6 and -1. Write an equation in general form.

Information given:

Point: (1, -7)

X-intercepts:

  • X1 = -6
  • X2 = -1

By looking at the information, we cannot use the standard form because we are not given the vertex.

We therefore have to use factored form as we have x1 and x2 as well as a point.

Factored form: y=a(x-x1)(x-x2)

Replace

  • y = -7
  • x = 1
  • x1: -6
  • x2 = -1
-7=a(1+6)(1+1)

When x1 is replaced with -6 and when x2 is replaced with -1, the sign changes to positive because there are two negatives.

Next is to isolate a.

  • -7=a(7)(2)
  • -7=14a
  • \frac{-7}{14}=\frac{14}{14}
  • -\frac{1}{2}=a

Now that you have a, put it in the equation. Remember, they want it in general form.

  • y=-\frac{1}{2}(x+6)(x+1)
  • y=-\frac{1}{2}(x^2+7x+6)
  • y=-\frac{1}{2}x^2-\frac{7}{2}-3