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Lesechka [4]
3 years ago
12

Write the quadratic equation y= x^2 -6x = 7 in vertex form.

Mathematics
1 answer:
lapo4ka [179]3 years ago
7 0

Answer:

(x - 3)² - 16 = 0

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Subtract 7 from both sides

x² - 6x + 7 = 0 ← in standard form

with a = 1, b = - 6

Given a quadratic in standard form then the x- coordinate of the vertex is

x_{vertex} = - \frac{b}{2a} = - \frac{-6}{2} = 3

Substitute x = 3 into the equation for y

y = 3² - 6(3) - 7 = 9 - 18 - 7 = - 16 ⇒ (h, k) = (3, - 16)

y = (x - 3)² - 16 = 0

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f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}

slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1

b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}

=\lim\limits_{x\to\pm\infty}\dfrac{9-\frac{3}{x}}{1-\frac{7}{x}}=\dfrac{9}{1}=9\\\\\boxed{y=1x+9}

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