Write x^2 + 4x +5 in the form of (x + a)^2 + b where a and b are integers

Mathematics

1 answer:

Alisiya [41]2 years ago

6 0

Answer:

$(x+2)^2 + 1 \ , \ where \ a = 2 \ , \ b = 1$

Step-by-step explanation:

$(x + 2)^2 = x^2 + 4x + 4\\\\(x+2)^2 + 1 = x^2 + 4x + 5$

$Therefore,$

$x^2 + 4x + 5 = (x+2)^2 + 1$

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The equation of line c is y–5=

Paraphin [41]

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the "c" equation

$y-5=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{3}{5}}(x-7)\qquad \impliedby \begin{array}{|c|ll}\cline{1-1}\textit{point-slope form}\\\cline{1-1}\\y-y_1=m(x-x_1)\\\\\cline{1-1}\end{array}$

so line "d" is a line with a slope of 3/5 and passes through (1,-1)

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-1})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{3}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{3}{5}}(x-\stackrel{x_1}{1})\implies y+1=\cfrac{3}{5}(x-1) \\\\\\ y+1=\cfrac{3}{5}x-\cfrac{3}{5}\implies y=\cfrac{3}{5}x-\cfrac{3}{5}-1\implies y=\cfrac{3}{5}x-\cfrac{8}{5}$