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# Ib st sl quadratic functions_test

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### Ib st sl quadratic functions_test

1. 1. Quadratic Functions <ul><li>IB Studies, Syllabus ref 4.3 </li></ul><ul><li>Lesson 1 </li></ul>
2. 2. Let us recall: <ul><li>A quadratic function is just another name for a parabola. </li></ul><ul><li>The curve is symmetrical about an axis of symmetry . </li></ul><ul><li>The curve has a turning point or vertex . </li></ul>Let’s start today by looking at the axis of symmetry .
3. 3. <ul><li>A parabola has a vertical axis of symmetry which passes through the vertex. </li></ul><ul><li>This means it has an equation in the form of a vertical line, that is _____ where ‘a’ is a constant. </li></ul><ul><li>The easiest way to find the axis of symmetry (AoS) is to graph the parabola, find the x-intercepts and then work out the x value in the middle of the 2 intercepts. </li></ul><ul><li>Alternatively if you are already given the vertex then you can read off the AoS equation directly from the coordinates of the vertex. </li></ul><ul><li>Lastly if you are required to find the coordinates of the vertex and you have the AoS you can just substitute the AoS value for ‘x’ in to the equation of the parabola. </li></ul>Axis of Symmetry (AoS) Let’s look at an example of each of these cases
4. 4. <ul><li>Firstly realise that the equation is in factored form and therefore we can read the x-intercepts straight from the equation. It cuts the x-axis at 2 and -4. </li></ul><ul><li>To place it exactly we should also find the y-intercept, by expanding just the constants i.e. -2 x 4 = -8 we have the y-intercept. </li></ul><ul><li>Now by sketching the function we can see that the axis of symmetry is at x=-1. Since half it is half way between 2 and -4. i.e. (-4+2)/2=-1 </li></ul><ul><li>Finally by substituting x=-1 in to the original equation we get y=(-3)(3)=-9. Hence the coordinates of the vertex is (-1,-9). </li></ul>Example 1 - Sketch the graph of and find the equation of the axis of symmetry and the coordinates of the vertex. and find the equation of the axis of symmetry and the coordinates of the vertex.
5. 5. <ul><li>Firstly you could actually do this without sketching but since the questions asks you to sketch you MUST sketch it. </li></ul><ul><li>To place it exactly we already have enough information, namely all the axis intercepts. </li></ul><ul><li>From the sketch it could simply be counted as half way between the x-intercepts or alternatively use the average technique: i.e. (-3+5)/2=+1 </li></ul><ul><li>Thus the equation of the AoS is x=1 </li></ul>Example 2 - Sketch the parabola which has x intercepts -3 and 5, and a y-intercept of -2. Find the equation of the axis of symmetry. Find the equation of the axis of symmetry.
6. 6. Now complete these in your books: Click here when you’ve completed them and marked them all! You only have to do the first column here!
7. 7. Last ones for today... Click here when you’ve completed them and marked them all!
8. 8. Well done! I’m impressed!