1 Investigating the Quadratic Function | Type I
(a)
(b)
(c)
1. What I notice about the three graphs is that they are all the same exact shape, but in different locations, specifically translated up or down according to the constant added at the end of the equation. I can generalize, based on these graphs, that adding a positive (negative) constant k onto the function x2 will translate up (down) that function x2 k units.
(a)
(b)
(c)
2. What I notice about the three graphs is that they are again all the same exact shape, but in different locations, this time specifically translated left or right according to the expression replacing x in each equation. I can generalize, based on these graphs, that replacing x in a function x2 with the expression (x-h) where h is positive (negative) will translate the function x2 to the right (left) h units.
3. Based on the information above, I would expect that the vertex would be at (4, 5). The equation y=x2 has its vertex at (0, 0). Using the information gathered above, the translation values (h, k) are (4, 5). Thus, the whole function would be moved to the right 4 units and up 5. Therefore, the original vertex (0, 0) would become (4, 5).
To express the expression x 2 - 10x + 25 in the form (x - h) 2, it is necessary to complete the square. This is simple because x2-10x+25 is a perfect square.
x 2 - 10x + 25 = (x-5)2
(b) Express x2 - 10x + 32 in the form (x-h)2 + g .
To express the expression x 2 - 10x + 32 in the form (x - h) 2 + g, it is necessary to complete the square. This is slightly more difficult because the expression is not a perfect square. However, it is equal to a perfect square plus a constant g.
x 2 - 10x + 32
x 2 - 10x + (25 + 7)
(x 2 - 10x + 25) + 7
(x-5) 2 +7
Thus, x 2 - 10x + 32 = (x-5) 2 +7 .
(c) Repeat this procedure with some examples of your own.
Using the examples (i) x2 +8x + 16, (ii) x2 + 8x + 19, and (iii) x2 - 5x + 8.5 .
(i) The expression x2 + 8x +16 is a perfect square, equal to (x + 4) 2 .
(ii) x2 + 8x + 19 is not a perfect square, but like x 2 - 10x + 32, it is a perfect square plus a constant.
x2 + 8x + 19
x2 + 8x + (16 + 3)
(x2 + 8x + 16) +3
(x + 4) 2 + 3 = x2 + 8x + 19
(iii) The expression x2 - 5x + 8.5 is not a perfect square, either. Some manipulation will be necessary. The middle term, of degree 1, determines the constant h of (x-h)2. Therefore, h will equal 5/2 . The square of (x-h), (x- 5/2)2, is x2 -5x + 25/4, which is x2 -5x + 6.25. Adding the necessary 2.25 to make it equal to the original expression, the expression in the x2 + bx + c is (x- 5/2)2 + 2.25 .
x2 + 5x + 8.5 = (x- 5/2)2 + 2.25
(d) Describe a method of writing the quadratic expression x2 + bx +c in the form (x-h)2 + g .
It is necessary to take into account that not all numbers are perfect squares when solving this problem. Because (a+b)2 is a x2 + 2*a*b + b x2, a perfect square x2 + bx + c would simplify to (x+b/2)2. To account for the difference of c and b/2 in nonperfect squares, the constant (c - (b/2)2) should be added to the expression (x + b/2)2. Thus, to write an expression x2 + bx + c in the form x2 + bx + c, one uses the formula (x + b/2)2 + (c - (b/2)2).
5. Per the information from questions 1 and 2, the graph of y = (x-h) 2 + g will be congruent with the graph of y = x2, but translated to the right h units (or left if h is negative) and up g units (or down if g is negative). The simplest explanation is that if f(x) = y, then f(x) + g = y + g, raising or lowering y in accordance with the sign of g. Likewise, at a given x value x0 of a function f(x), the same y value would be obtained at x0-h if the function f(x-h) is being used.