X0 2 4 6 8 f(x) 15 15.5 17 19.5 23 write the quadratic function that corresponds to the given table of values in standard form, f(x) = ax2 + bx + c, where a, b and c are constants. f(x)=ax^2+bx+c

Answer with Step-by-step explanation:

We are given that:

f(x)=ax²+bx+c

also at x=0 f(x)=15

Putting x=0 in f(x)

a×0+b×0+c=15

⇒ c=15

now, f(2)=15.5

⇒ a×2²+b×2+c=15.5

⇒ 4a+2b+15=15.5

⇒ 4a+2b=0.5  (1)

f(4)=17

⇒ a×4²+b×4+c=17

⇒ 16a+4b+15=17

⇒ 16a+4b=2     (2)

equation (2)- 2×equation(1)

⇒ 16a+4b-8a-4b=2-2×0.5

⇒ 8a=2-1

⇒ 8a=1

⇒ a=1/8

⇒ a=0.125

Putting value of a and c in equation (1)

⇒ 4×0.125+2b=0.5

⇒ 0.5+2b=0.5

⇒ 2b=0

⇒ b=0

Hence

a=0.125

b=0

c=15

and f(x)=0.125x²+15

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)