Heheboi0728idn
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[tex]y = x {}^{2} - 9x + 14[/tex]

Given that y can be written in the form
[tex](x + a) {}^{2} + b[/tex]
, where a and b are constants, find the value of a and the value of b.​
pls help i rly need this Pls pls pls pls

Sagot :

Answer:

a = - [tex]\frac{9}{2}[/tex] , b = - [tex]\frac{25}{4}[/tex]

Step-by-step explanation:

To obtain the required form use the method of completing the square

add/ subtract ( half the coefficient of the x- term)² to x² - 9x

y = x² + 2(- [tex]\frac{9}{2}[/tex] )x + [tex]\frac{81}{4}[/tex] - [tex]\frac{81}{4}[/tex] + 14

  = (x - [tex]\frac{9}{2}[/tex] )² - [tex]\frac{81}{4}[/tex] + [tex]\frac{56}{4}[/tex]

  = (x - [tex]\frac{9}{2}[/tex] )²- [tex]\frac{25}{4}[/tex] ← in the form (x + a)² + b

with a = - [tex]\frac{9}{2}[/tex] and b = - [tex]\frac{25}{4}[/tex]

VirαtKσhliidn

Answer:

a = 4.5

b = -6.25

Step-by-step explanation:

The given equation to us is ,

[tex]\implies y = x {}^{2} - 9x + 14[/tex]

And its given that it can we written in the form of ,

[tex]\implies (x + a) {}^{2} + b[/tex]

Where ,

  • a and b are constants .

Therefore ,

[tex]\implies y = x^2 -9x + 14 [/tex]

Multiplying 9x by 2/2 ,we have ,

[tex]\implies y = x^2 -\dfrac{2}{2}\times 9 x + 14 [/tex]

Adding and subtracting (9/2)² ,

[tex]\implies y = \bigg\{ x^2 -\dfrac{2}{2}\times 9 x + \bigg(\dfrac{9}{2}\bigg)^2 \bigg\} +14 -\bigg(\dfrac{9}{2}\bigg)^2[/tex]

Therefore , we can write it as ,

[tex]\implies y = \bigg( x + \dfrac{9}{2}\bigg)^2 + 14 - 20.25 \\\\\implies \underline{\underline{ y = \bigg( x + \dfrac{9}{2}\bigg)^2 - 6.25 }}[/tex]

Hence the value of a is 9/2 and b is -6.25 .

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