Mathematics, 09.07.2019 09:00 Leii691

How do you rewrite : y=x^2-5x+9 in standard form using complying the square? i keep getting to : x^2-5x-25/4 but i’m not allowed to use a calculator so how do i factor that? ?

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Answer from: mehokkj

I presume with this you meant to say you want this equation in general form, also known as vertex form, which is y = a(x - h)² + k, instead of standard form, which is y = ax² + bx + c, which is what the original equation is already in.

So with completing the square, you first have to isolate the x-terms. To do this, subtract 9 on both sides of the equation:

y-9=x^2-5x

Next, we want to make the right side of the equation a perfect square. To find the constant of this soon-to-be perfect square, divide the x-coefficient by 2 and square that quotient. In this case:

$-5\div 2 = -\frac{5}{2}\\\\(-\frac{5}{2})^2=-\frac{5}{2}\times-\frac{5}{2}=\frac{25}{4}$

Now add 25/4 on both sides of the equation:

$y-9+\frac{25}{4}=x^2-5x+\frac{25}{4}$

Now to combine -9/1 and 25/4, they must have the same denominator, and to find it you must find the LCD, or lowest common denominator, of 1 and 4. To find it, list the multiples of both and the lowest one they share is their LCD. In this case, the LCD is 4. Multiply both sides of -9/1 by 4/4 and then add the numerators up:

$-\frac{9}{1}\times \frac{4}{4}=-\frac{36}{4}\\\\-\frac{36}{4}+\frac{25}{4}=-\frac{11}{4}\\\\y-\frac{11}{4}=x^2-5x+\frac{25}{4}$

Next, we need to factor the right side of the equation. Using this format of a^2-2ab+b^2=(a-b)^2 , we can factor it as:

$y-\frac{11}{4}=(x-\frac{5}{2})^2$

Now lastly, add both sides by 11/4, and your final answer will be $y=(x-\frac{5}{2})^2+\frac{11}{4}$

Answer from: Quest

0 is its x and y coordinate

Line a b represents a proportional relationship. point a lies at 6, 4 as shown on the graph below :

Answer from: Quest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

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