label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

f(x) = −x2 − 3x + 3
(a) Express the quadratic function in standard form.
 f(x)

(c) Find its maximum or minimum value.

f(x) =
Jul 8th, 2015

The standard form of a quadratic function is

$\\ f(x)=a(x-h)^2+k\\ \\$

where (h, k) is the vertex of the parabola.

$\\ f(x)=-x^2-3x+3\\ \\$

f(x) can be expressed in standard form by completing the square.

First make the coefficient of x^2 1 by taking -1 common

$f(x)=-(x^2+3x-3)$

Take half of coefficient of x which gives 3/2

and square it which gives (3/2)^2 = 9/4

Add and subtract 9/4 from the equation inside the bracket.

$\\ f(x)=-\bigg(x^2+3x-3+\frac{9}{4}-\frac{9}{4}\bigg)\\ \\ f(x)=-\bigg(\bigg[x^2+3x+\frac{9}{4}\bigg]-3-\frac{9}{4}\bigg)\\ \\ f(x)=-\bigg(\bigg[x^2+3x+\bigg(\frac{3}{2}\bigg)^2\bigg]+\frac{-12-9}{4}\bigg)\\ \\ f(x)=-\bigg(\bigg[x+\frac{3}{2}\bigg]^2-\frac{21}{4}\bigg)\\ \\ f(x)=-\bigg(x+\frac{3}{2}\bigg)^2+\frac{21}{4}\\$

This is the standard form of the quadratic function.

Since a = -1 < 0, the parabola opens downwards.

Hence the vertex is the highest point on the parabola.

Vertex = (h, k) = (-3/2, 21/4)

So, maximum value = 21/4

a) $f(x)=-\bigg(x+\frac{3}{2}\bigg)^2+\frac{21}{4}\\$

b) Maximum value is f(x) = 21/4

Jul 8th, 2015

...
Jul 8th, 2015
...
Jul 8th, 2015
Sep 21st, 2017
check_circle