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V125BC [204]
4 years ago
9

Erin was completing the square of the quadratic function in order to find the extreme value. What is the next step in the proces

s, and what is the extreme value?
Mathematics
1 answer:
Lady bird [3.3K]4 years ago
7 0

Step-by-step explanation:

ax^2+bx+c=0

a=leading term

ok so if the leading term is positive then opens up and has a <u>min</u>

if leading term is negative then opens down and has a <u>max</u>

leading term is positive

1x^2+8x

it has a min

to complete the square, move c aside take 1/2 of b and square it

b=8

8/2=4

4^2=16

now add that to both sides

x^2+8x+16+6=0+16

factor perfect square

(x+4)^2+6=16

subtract 6

(x+4)^2=10

subtract 10

(x+4)^2-10=0

vertex aka min or max is (h,k) when ou have

y=a(x-h)+k

h=-4

k=-10

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Step-by-step explanation:

Y= -x-2;(2,4)


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Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

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