Week 8 – Analyzing the Standard Form

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This week in Pre-calculus 11 I learned  a lot about graphing quadratic functions and how to analyze them. Something that stuck with me was how to analyze quadratic functions using the standard form: y=a{(x-p)}^2+q.

Things to know:

  • General Form: y=ax^2+bx+c, when using the general form, c tells you what the y-intercept is, and a tells if the parabola will open up or down (if a is negative it will open down, if a is positive it will open up) as well as if the parabola will be stretched or compressed.
  • Standard Form: y=a{(x-p)}^2+q, when using the standard form, a will tell you if the parabola will open up or down, if the parabola will be stretched or compress and the scale. In the equation, p will tell you the horizontal translation and q will tell you the vertical translation. The vertex will always be represented by (p,q).
  • Scale: The scale of a parabola can be found by looking at the coefficient in front of the x^2. If the coefficient in front is equal to one then the parabola will have a scale of 1-3-5 (to check this: Start at vertex go up 1 over 1, up 3 over one, up 5 over one etc. if it matches then the parabola has a scale of 1-3-5), having a scale of 1-3-5 also means that it is congruent to the parent function y=x^2, it can also have a scale of 2,6,10.
  • Vertex: The highest or lowest point of the parabola.
  • Axis of Symmetry: intersects the parabola at the vertex. Splits the parabola perfectly in half.
  • Minimum & Maximum: used to indicate if the vertex is at the highest or lowest possible point. If the parabola is opened up it will be at its lowest height (minimum), if the parabola is opened down the parabola is at its highest point (maximum).
  • Domain: All possible values of x, will always be x\in\Re
  • Range: All possible values of y

 

Example:

  • Firstly since our equation is written in the proper format (y=a{(x-p)}^2+q) we should write down what we know and put it into the formula. We know: a =1, p =2, and q =6 [ ].
  • To start off it is best to find what the vertex is since it tells you a lot about a quadratic function. To find the quadratic function we use what we know: (p,q) –> (-2,6). If we know that the vertex is (-2,6) then we know that the axis of symmetry is: x= -2 (the axis of symmetry is located where the vertex is). If we then take a look at the value of a we know that the parabola of this equation will open up and have a scale of 1-3-5 (congruent to parent function: y=x^2). We know that it will open up because the value of a is positive and we know it will have a scale of 1-3-5 because the value of a is 1. If we know that the parabola opens up then we know that it is at its lowest point (minimum) where y =6. Finally based on what we know we can write down the domain and range. Since this is a parabola the domain will always be  x\in\Re. By looking at the function we know that the range will be y≥6 because the lowest that the parabola goes on the y axis is 6, and since it is opening up, all possible values for y would be greater than or equal to 6.

Example:

  • Firstly since our equation is written in the proper format (y=a{(x-p)}^2+q) we should write down what we know and put it into the formula. We know: a =-2, p =-3, and q =5 [].
  • Since we have put what we know into the formula we can write what the vertex is: (p,q) –> (3,5)
  • If the vertex is (3,5) then we know that the Axis of symmetry is x = 3
  • Since the value of a is -2 we know that the parabola will open downwards and will be at its highest point (maximum). We also know that it will have a scale of 2-6-10 (because it has a value of 2 and not 1, you double the scale of 1-3-5 –> 2-6-10) and not congruent to the parent function.
  • We also know that the domain is x\in\Re and the range is y≤5 (highest point on the y axis is 5, and the parabola is facing downwards, there would be no greater value than 5).

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