Standard form to vertex form??

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Standard form to vertex form??

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Answer 1 · 2018-03-29T03:22:03

Complete the square

Explanation:

We want to go from y intercept form $f (x) = a x^{2} + b x + c$ $f(x)=ax^2+bx+c$ into vertex form $f (x) = a {(x - b)}^{2} + c$ $f(x)=a(x-b)^2+c$

So take the example of
$f (x) = 3 x^{2} + 5 x + 2$ $f(x)=3x^2+5x+2$

We need to factorise the co-efficient out from the $x^{2}$ $x^2$ and separate the $a x^{2} + b x$ $ax^2+bx$ from the $c$ $c$ so you can act upon them separately

$f (x) = 3 (x^{2} + \frac{5}{3} x) + 2$ $f(x)=3(x^2+5/3x) +2$

We want to follow this rule
$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2=(a+b)^2$
or
$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$

We know that the $a^{2} = x^{2}$ $a^2=x^2$ and
$2 a b = \frac{5}{3} x$ $2ab=5/3x$ so $2 b = \frac{5}{3}$ $2b=5/3$

So we just need $b^{2}$ $b^2$ and then we can collapse it down to ${(a + b)}^{2}$ $(a+b)^2$

so $2 b = \frac{5}{3}$ $2b=5/3$ so $b = \frac{5}{6}$ $b=5/6$ so $b^{2} = {(\frac{5}{6})}^{2}$ $b^2=(5/6)^2$

Now we can add the $b^{2}$ $b^2$ term into the equation remembering that the net sum of any additions to any equation/expression must be zero)

$f (x) = 3 (x^{2} + \frac{5}{3} x + {(\frac{5}{6})}^{2}) + 2 - 3 {(\frac{5}{6})}^{2}$ $f(x)=3(x^2+5/3 x+(5/6)^2)+2-3(5/6)^2$

Now we want to make the $a^{2} + 2 a b + b^{2}$ $a^2+2ab+b^2$ into ${(a + b)}^{2}$ $(a+b)^2$ so follow the same process as above

$f (x) = 3 {(x + \frac{5}{6})}^{2} + \frac{72}{36} - 3 (\frac{25}{36})$ $f(x)=3(x+5/6)^2+72/36-3(25/36)$

Simply the equation
$f (x) = 3 {(x + \frac{5}{6})}^{2} - \frac{3}{36}$ $f(x)=3(x+5/6)^2-3/36$

Now we have the result in standard form

Answer 2 · 2018-03-29T04:56:03

General vertex form of a quadratic function:
$f (x) = a {(x + \frac{b}{2 a})}^{2} + f (- \frac{b}{2 a})$ $f(x) = a(x + b/(2a))^2 + f(-b/(2a))$
In this formula,
$(- \frac{b}{2 a})$ $(-b/(2a))$ is the x-coordinate of the vertex
$f (- \frac{b}{2 a})$ $f(-b/(2a))$ is the y-coordinate of the vertex.
To proceed, first find $x = - \frac{b}{2 a}$ $x = -b/(2a)$ .
Next, find $f (- \frac{b}{2 a})$ $f(-b/(2a))$
Example: Transform to vertex form -->
$f (x) = x^{2} + 2 x - 15$ $f(x) = x^2 + 2x - 15$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{2}{2} = - 1$ $x = - b/(2a) = -2/2 = - 1$
y-coordinate of vertex:
$f (- \frac{b}{2 a}) = f (- 1) = 1 - 2 - 15 = - 16$ $f(-b/(2a)) = f(-1) = 1 - 2 - 15 = - 16$
Vertex form:
$f (x) = {(x + 1)}^{2} - 16$ $f(x) = (x + 1)^2 - 16$

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Standard form to vertex form??

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