Question

Complete the solutions in transforming the quadratic functions from quadratic function from y = ax2 + bx + c to y = a(x − h)2 + k

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{y=ax^2+bx+c}$

$\sf{\implies\,y=a\left(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\right)}$

$\sf{\implies\,y=a\left\{x^2+2\cdot\dfrac{b}{2a}\cdot\,x+\left(\dfrac{b}{2a}\right)^2+\dfrac{c}{a}-\left(\dfrac{b}{2a}\right)^2\right\}}$

$\sf{\implies\,y=a\left\{x^2+2\cdot\dfrac{b}{2a}\cdot\,x+\left(\dfrac{b}{2a}\right)^2\right\}+a\left\{\dfrac{c}{a}-\left(\dfrac{b}{2a}\right)^2\right\}}$

$\sf{\implies\,y=a\left(x+\dfrac{b}{2a}\right)^2+a\left(\dfrac{c}{a}-\dfrac{b^2}{4a^2}\right)}$

$\sf{\implies\,y=a\left\{x-\left(-\dfrac{b}{2a}\right)\right\}^2+a\left(\dfrac{4ac-b^2}{4a^2}\right)}$

$\sf{\implies\,y=a\left\{x-\left(-\dfrac{b}{2a}\right)\right\}^2+\left(\dfrac{4ac-b^2}{4a}\right)}$

Answer 2

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:y = {ax}^{2} + bx + c$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{1}{4a} \bigg(4 {a}^{2} {x}^{2} + 4abx + 4ac\bigg)$

$\rm :\longmapsto\:y = \dfrac{1}{4a} \bigg({(2ax)}^{2} + 2(2ax)b + 4ac\bigg)$

$\rm :\longmapsto\:y = \dfrac{1}{4a} \bigg({(2ax)}^{2} + 2(2ax)b + {b}^{2} - {b}^{2} + 4ac\bigg)$

$\rm :\longmapsto\:y = \dfrac{1}{4a} \bigg([{(2ax)}^{2} + 2(2ax)b + {b}^{2}] - {b}^{2} + 4ac\bigg)$

$\rm :\longmapsto\:y = \dfrac{1}{4a}\bigg( {(2ax + b)}^{2} + 4ac - {b}^{2} \bigg)$

$\rm :\longmapsto\:y = \dfrac{1}{4a} {(2ax + b)}^{2} + \dfrac{4ac - {b}^{2} }{4a}$

$\rm :\longmapsto\:y = \dfrac{1}{4a} {\bigg[2a\bigg(x + \dfrac{b}{2a}\bigg) \bigg]}^{2} + \dfrac{4ac - {b}^{2} }{4a}$

$\rm :\longmapsto\:y = \dfrac{ {4a}^{2} }{4a} {\bigg[\bigg(x + \dfrac{b}{2a}\bigg) \bigg]}^{2} + \dfrac{4ac - {b}^{2} }{4a}$

$\rm :\longmapsto\:y = a{\bigg[\bigg(x + \dfrac{b}{2a}\bigg) \bigg]}^{2} + \dfrac{4ac - {b}^{2} }{4a}$

$\rm :\longmapsto\:y = a{\bigg[x - \bigg( - \dfrac{b}{2a}\bigg) \bigg]}^{2} + \dfrac{4ac - {b}^{2} }{4a}$

is the required transformation.

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EXPLORE MORE

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac