Slightly different result

Using what we learned from the first two forms we can work out where the vertex is going to be before we see it on a graph. We established that any negative numbers added directly to x would move the vertex that value to the right, we also worked out that adding a value to xi?? would move the vertex up that amount. So now we combine the two. The number added directly to x is -4, so it will move 4 units to the right, combined with moving up 5 units would give us the coordinate (4, 5). This is where the vertex is located on the graph.

So when the form ‘y = a(x + b)i?? + c’ is given the graph will move ‘b’ units to the right (left if it is positive) and c units up (down if it is negative) which is where the vertex will be located for that graph. The basic quadratic equation ‘y = axi?? + bx + c’ can be changed into turning point form to make it easier to find the vertex and the shape of the parabola. Eg: If we have the equation ‘xi?? – 10x + 25’ and we want to change it to the form (x – h)i?? , a perfect square, this method can be used: Divide 10 by 2, this will give us -5.

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Now plug this into the equation so that the x and the 5 are in brackets and all squared so the new equation will look like this: y = (x -5)i?? + 25 But now that 25 is being added in the brackets (-5i?? = 25) 25 has to be subtracted as well so now the equation will look like this: y = (x – 5)i?? +25 – 25 y = (x – 5)i?? The value that was being added to xi?? as a whole was eliminated so this answer is now in a perfect square and taking the square root of the right side would give us values for x. Similar to the previous calculation trying to work out y = xi?? – 10x + 32 into turning point form would give us slightly different results.

Notice that the value for ‘c’ in the form has changed from 25 to 32. The only difference is that the 25 would not be eliminated this time, as the value has now changed to 32 and 32 – 25 = 7 so now the answer will be y = (x – 5)i?? + 7. The only difference being the added 7, which will give us a y-value for the graph. Here is a graph of the two parabolas. The red curve is the first one and the blue curve is the second equation. H y = (x – 5)i?? y = (x – 5)i?? + 7 Notice how the line moves from 0 to 7 into the second curve as the 7 is added to the equation. It clarifies even more how adding a value to xi?? moves the parabola up that amount.