Question

Which equation is y = –6x^2 + 3x + 2 rewritten in vertex form? y = negative 6 (x minus 1) squared + 8 y = negative 6 (x + one-fourth) squared + thirteen-eighths y = negative 6 (x minus one-fourth) squared + nineteen-eighths y = negative 6 (x minus one-half) squared + seven-halves

Answer 1

Answer:

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

Step-by-step explanation:

Given:

$y = -6x^2 + 3x + 2$

Required

Rewrite in vertex form

The vertex form of an equation is in form of: $y = a(x - h)^2+ k$

Solving: $y = -6x^2 + 3x + 2$

Subtract 2 from both sides

$y - 2 = -6x^2 + 3x + 2 - 2$

$y - 2 = -6x^2 + 3x$

Factorize expression on the right hand side by dividing through by the coefficient of x²

$y - 2 = -6(x^2 + \frac{3x}{-6})$

$y - 2 = -6(x^2 - \frac{3x}{6})$

$y - 2 = -6(x^2 - \frac{x}{2})$

Get a perfect square of coefficient of x; then add to both sides

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Rough work

The coefficient of x is $\frac{-1}{2}$

It's square is $(\frac{-1}{2})^2 = \frac{1}{4}$

Adding inside the bracket of $-6(x^2 - \frac{x}{2})$ to give: $-6(x^2 - \frac{x}{2} + \frac{1}{4})$

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: $-6(\frac{1}{4})$

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The expression becomes

$y - 2 -6(\frac{1}{4})= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

Factorize the expression on the right hand side

$y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2$

$y - (2 +\frac{3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{4+3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{7}{2})= -6(x - \frac{1}{2})^2$

$y +\frac{7}{2} = -6(x - \frac{1}{2})^2$

Make y the subject of formula

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

Solved

Answer 2

Answer

Y= -6( x - 1/2)^2 + 7/2

Step-by-step explanation:

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