Which of the following functions is shown in the graph below?
y = (x ­- 1)^2 + 2
y = -2(x -­ 1)^2 + 2
y = 2(x + 1)^2 + 2
y = (x + 1)^2 + 2

y = -2(x -­ 1)² + 2

Step-by-step explanation:

recall that the vertex form of a quadratic equation is :

y = a(x - h)² + k, where (h, k) is the coordinate of the vertex (i.e maxima point)

from the graph we can see that the vertex is at x = 1, y = 2

hence the answers that are valid would have h = 1 and k = 2

right away by observing the answers, we can see that the last 2 choices have h = (-1) and so these are NOT the answers.

To decide between the first 2 choices, we observe from the given graph, that when x=0, y=0.

The only choice which satisfies this is the 2nd option

Proof:

for y = -2(x -­ 1)² + 2, when x = 0,

y = -2(0 -­ 1)² + 2

y = -2(-­ 1)² + 2

y = -2(1) + 2 = -2 + 2 = 0 (Proven to be valid)

Sanity check: check the first choice

y = (x ­- 1)² + 2

when x = 0,

y = (0- 1)² + 2

y = 1 + 2 ≠ 0 (not the answer)

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.