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So it is the product of
And 2,5 and 11 are prime numbers, so there you go :)
Part A:
Given parabola (1) to be
, and parabola (2) to be 
Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).
Therefore, parabola (1) is wider than parabola (2).
Part B:
A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.
Given parabola (1) to be, and parabola (2) to be
Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).
Part C:
For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).
Given parabola (1) to be, and parabola (2) to be
Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.
Therefore, the parabola that would be furthest left on the x-axis is parabola (1).
Part D:
For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).
Given parabola (1) to be, and parabola (2) to be
Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.
Therefore, the parabola that would be highest on the y-axis is parabola (2).
Given parabola (1) to be
Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).
Therefore, parabola (1) is wider than parabola (2).
Part B:
A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.
Given parabola (1) to be
Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).
Part C:
For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).
Given parabola (1) to be
Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.
Therefore, the parabola that would be furthest left on the x-axis is parabola (1).
Part D:
For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).
Given parabola (1) to be
Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.
Therefore, the parabola that would be highest on the y-axis is parabola (2).