Dezembro 23, 2019

por Aeriela

Como você reescreve $y = {(x + 3)}^{2} + {(x + 4)}^{2}$ $y = (x + 3) ^ 2 + (x + 4) ^ 2$ na forma de vértice?

Responda:

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

Explicação:

The equation of a parabola in $vertex form$ $color(blue)"vertex form"$ is.

$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ ∣ ∣ ∣ \frac{2}{2} y = a {(x - h)}^{2} + k \frac{2}{2} ∣ ∣ ∣ - ------------------- -$ $color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))$
where ( h , k ) are the coordinates of the vertex and a is a constant.

$Expand and simplify y$ $"Expand and simplify y"$

$y = x^{2} + 6 x + 9 + x^{2} + 8 x + 16$ $y=x^2+6x+9+x^2+8x+16$

$y = 2 x^{2} + 14 x + 25$ $color(white)(y)=2x^2+14x+25$

$using the method of completing the square$ $"using the method of "color(blue)"completing the square"$

$y = 2 (x^{2} + 7 x) + 25 \leftarrow coefficient of x^{2} term is unity$ $y=2(x^2+7x)+25larr" coefficient of " x^2" term is unity"$

add ${(\frac{1}{2} coefficient of x-term)}^{2} to x^{2} + 7 x$ $(1/2"coefficient of x-term")^2" to " x^2+7x$

$we must also subtract this value$ $"we must also subtract this value"$

$that is add/subtract {(\frac{7}{2})}^{2} = \frac{49}{4}$ $"that is add/subtract " (7/2)^2=49/4$

$y = 2 (x^{2} + 7 x + \frac{49}{4} - \frac{49}{4}) + 25$ $y=2(x^2+7xcolor(red)(+49/4)color(red)(-49/4))+25$

$y = 2 {(x + \frac{7}{2})}^{2} - \frac{49}{2} + 25$ $y=2(x+7/2)^2-49/2+25$

$\Rightarrow y = 2 {(x + \frac{7}{2})}^{2} + \frac{1}{2} \leftarrow in vertex form$ $rArry=2(x+7/2)^2+1/2larrcolor(red)" in vertex form"$