INCORRECT ANSWERS:
1.The figure above shows the graph of a quadratic function h whose maximum value is h(2). If h(a)=0 which of the following could be the value of a?
*-1
*0
*2
*3
*4
Well i put 2 in the spot of h and got 2a=0 so i thought the answer was 0. How do i do this?
2. If k is a positive integer, which hof the following myst represent an even integer that is twice the value of an odd integer?
*2k
*2k+3
*2k+4
*4k+2
*4k+1
I thought it would be 2k+3 because i substituted an odd integer into k and that worked. But the answer was 4k +2?
1 comment:
1) h refers to the name of the function, not any particular variable; this function is not described other than that it is quadratic, and the number in parentheses is the x-value. 'a' would therefore be one of the two values of x for which the corresponding y-value is 0, i.e. the x-intercepts.
Parabolas are symmetrical. The parabola does not hit the x-axis on the left side until it clears the y-axis, so it must be more than two steps to the left of the maximum (i.e. the center).
0 can't be it, because the curve does not pass through the origin. 2 is the maximum. 3 and 4 can't be the answers, because if the curve took only 1 or two steps to the right to hit the x-axis, it would require the same on the left, meaning it would hit the x-axis prior to hitting the y-axis.
This leaves only -1 as a possibility.
For the second, k could be either even or odd. To ensure that we get an odd number from it, we multiply by 2 (guaranteeing even) and add 1; this gives us 2k+1.
Twice the value of this integer would be 2(2k+1), which by the distributive property would be 4k+2.
2k+3, by contrast, must be an odd number. Regardless of what k is, 2k must be even. If you add 3, this gives us an odd number.
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