Question

NO LINKS!! URGENT HELP PLEASE!!!!

Find the equation of square root with a vertex at (1, -5) and passes through the (5, -7).

Answer 1

`f(x) = -√(x-1) - 5`

Explanation:

The standard form of a square root function is:

`f(x)=a√(x-h)+k`

where "a" is a scaling factor determining the vertical stretch or compression of the function, and (h, k) is the vertex.

Given the vertex is (1, -5), substitute h = 1 and k = -5 into the equation:

`f(x) = a√(x - 1) - 5`

Substitute the point (5, -7) that the function passes through to find the stretch factor "a":

`\begin{aligned}-7 &= a √(5 - 1) - 5\\-7 &= a √(4) - 5\\-7 &= 2a - 5\\-7 +5&= 2a - 5+5\\-2&= 2a\\(-2)/(2)&=(2a)/(2)\\-1&=a\\a &= -1\end{aligned}`

Substitute the value of "a" back into the equation:

`f(x) = -√(x-1) - 5`

Therefore, the equation of the square root function with a vertex at (1, -5) and passes through the point (5, -7) is:

`\boxed{f(x) = -√(x-1) - 5}`

Answer 2

`y = ( - 1)/(2) *√(x-1) - 5`

Explanation:

The standard equation of a square root function with vertex at point (h,k) is given by:

`y = a * √(x - h) + k`

where (h,k) is the vertex and "a" is a coefficient that determines the vertical stretch and direction of the square root function.

We know that the vertex of our square root function is (1, -5), so we can substitute those values into the equation to get:

y =
`a*√(x-1) - 5`

Now, we need to find the value of "a". To do that, we will use the fact that the square root function passes through point (5, -7). Substituting these coordinates into the equation, we get:

`-7 = a*\sqrt(5-1) - 5`

-2 = 4a

`a = ( - 1)/(2)`

Now that we know the value of "a", we can substitute it into our equation to get the final equation of the square root function:

`y = ( - 1)/(2) *√(x-1) - 5`

Therefore, the equation of the square root function with a vertex at (1, -5) and passing through (5, -7) is

`y = ( - 1)/(2) *√(x-1) - 5`