Aparabola has its vertex at (2, 1) and crosses the x-axis at (1,0) anu al l
the equations below has this parabola for its graph? (teks a2.4b-r)

ay=1-(x - 2)^2
c y = -1 - (x + 2)^2
dy= -1 - (x - 2)^2
b
y = 1 + (x + 2)^2

A

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (2, 1), thus

y = a(x - 2)² + 1

To find a substitute (1, 0) into the equation

0 = a(1 - 2)² + 1

0 = a + 1 ( subtract 1 from both sides )

a = - 1

Hence

y = - (x - 2)² + 1 or

y = 1 - (x - 2)² → A

5x + 5

step-by-step explanation:

you should always add the diagrams and graphs we cannot see. this particular question is kind of borderline because we can make a graph for you.

(f + g)x = -x^2 +3x + 5 + x^2 + 2x

(f + g)x = 5x + 5

the graph of z = 5x + 5 is shown below. always use desmos when in doubt.