Question

Which equation has a graph that lies entirely above the x-axis?

Answer 1

Answer:

• y = $(x-7)^{2}$ + 7

Step-by-step explanation:

The graph that will lie above the x-axis will be having only positive values throughout its domain,

So, We have to check which graph is having y > 0 for every x.

• y = -$(x+7)^{2}$ + 7

clearly y will be negative for many values of x , as coefficient of variable is negative.

• y = $(x-7)^{2}$ - 7

Put x = 0, you will get y as negative .

• y = $(x-7)^{2}$ + 7

Since, Square of anything will always be positive , and here constant term is also positive , so, It will always be positive .

Thus, it is having its graph always above x axis.

• y = x - 7

Put x = 0 , y = -7, which is negative

Answer 2

Answer:

Option C.

Step-by-step explanation:

The vertex form of a parabola is

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

where, a is constant, (h,k) is vertex.

If a<0, then it is a downward parabola and if a>0, then it is an upward parabola.  

A downward parabola never lies entirely above the x-axis.

First equation is

$y=-(x+7)^2+7$

It is a downward parabola and vertex is (-7,7).

Second equation is

$y=(x-7)^2-7$

It is an upward parabola and vertex is (7,-7).

Third equation is

$y=(x-7)^2+7$

It is an upward parabola and vertex is (7,7).

Fourth equation is

$y=(x-7)$

It is a linear equation with y-intercept -7.

Only equation 3 is an upward parabola whose vertex lies above the x-axis.

Hence the correct option is C.