Find the vertex of f(x) = 2x^2 +6x +10
a.) (3 , 0)
b.) (-1.5,4,5)
c.) (-3,4)
d.) (-1.5, 5.5)

Step-by-step explanation:

To find the vertex of 2x^2 + 6x + 10, you must complete the square.

(2x^2 + 6x          ) + 10 < start by separating constant from variable terms

2 (x^2 + 3x         ) + 10 < factor out the coefficient of first term

2 (x^2 + 3x + 2.25) + 10 - 4.5 <--- divide coefficient of x-term by 2, square result. then add inside parenthesis and subtract outside parenthesis (after multiplying it by 2 because we factored it out)

2 (x^2 + 3x + 2.25) + 10 - 4.5 <--- then simply factor trinomial and simplify outside constant terms

after factoring and simplifying, you should have 2 (x + 1.5)^2 + 5.5, which is vertex form.

you can find the vertex from vertex form, so the vertex is (-1.5, 5.5), or option D.

have a good day!

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

by definition, an arc subtended by a central angle of some given measure has the same measure. so .