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What is the range of the graph of #y = 5(x – 2)^2 + 7#?

Answer 1

#color(blue)(y in[7,oo)#

Notice #y=5(x-2)^2+7# is in the vertex form of a quadratic:

#y=a(x-h)^2+k#

Where:

#bba# is the coefficient of #x^2#, #bbh# is the axis of symmetry and #bbk# is the maximum/minimum value of the function.

If:

#a>0# then the parabola is of the form #uuu# and #k# is a minimum value.

For instance:

#5>0#

#k=7#

so #k# is a minimum value.

We now see what happens as #x->+-oo#:

as #x->oocolor(white)(88888)#, #5(x-2)^2+7->oo#

as #x->-oocolor(white)(888)#, #5(x-2)^2+7->oo#

Thus, in interval notation, the function's range is:

#y in[7,oo)#

This is confirmed by the graph of #y=5(x-2)^2+7#

plot{y=5(x-2)^2+7 [-10, 10, -5, 41.6]}

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Answer 2

Kennedy Adams

The given quadratic function is in vertex form (y = a(x - h)^2 + k), where ((h, k)) is the vertex of the parabola. The range of the graph is determined by the value of (a).

For (y = 5(x - 2)^2 + 7), the minimum value occurs at the vertex ((2, 7)), and since (a = 5) is positive, the parabola opens upwards. Therefore, the range is (y \geq 7).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.