Y=x^2-2x-5 vertex form

Hey there! I'm happy to help!

First, we'll want to find the vertex of this parabola.

Given the equation y=ax^2+bx+c, the formula for the x-coordinate of the vertex is always .

In our equation here, our a is 1 and our b is -2.

Now that we have our x-value of our vertex, we can plug it into our equation to find our y-coordinate.

y=1²-2(1)-5

y=1-2+5

y=-1+5

y=4

So, our vertex is at (1,4).

Vertex form is y=a(x-h)²+ k where (h,k) is our vertex. We know now that our vertex is (1,4) and we also know that our a is 1, so we can simply plug it in to this equation.

y=(x-1)²+4

Have a wonderful day and keep on learning! :D

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

d. 145*2 is 290 than the total angle is 360 then minus 290 then divide by two