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Help please!!!!!!!!!!!!

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Sagot :

Step-by-step explanation:

1.

the y-intercept comes automatically from the given form, as the y-intercept is the y value when x = 0.

so, in

x² - 12x + 46

the y-intercept is 46, or as point (0, 46).

now to the vertex form and vertex itself.

the vertex form for such a quadratic equation is

y = a(x - h)² + k

with (h, k) being the vertex.

since we have a plain x² term in the basic expression, "a" must be 1.

so, we have the simpler version

y = (x - h)² + k = x² - 2hx + h² + k

so, let's compare this with our basic expression, and we see

-2h = -12

h = 6

h² + k = 46

6² + k = 46

36 + k = 46

k = 10

so, the vertex form is

y = (x - 6)² + 10

and the vertex is (6, 10)

2.

the equating of a circle is

(x - x1)² + (y - y1)² = r²

with (x1, y1) being the center of the circle, and r being the radius.

so, the center is then (3, -3), and the radius is sqrt(36)=6.

3.

as explained in 2. this would be

(x - 4)² + (y + 9)² = 3² = 9

4.

x² + y² + 6x - 2y + 9 = 0

this has to turn into

(x - x1)² + (y - y1)² = r²

x² - 2x×x1 + x1² + y² - 2y×y1 + y1² = r²

x² - 2x×x1 + x1² + y² - 2y×y1 + y1² - r² = 0

"x² + 6x" suggests that for x1 of the first term

6x = - 2x×x1

x1 = -3

and "y² - 2y" suggests for y1 of the second term

-2y = -2y×y1

y1 = 1

that gives us for the constant terms and radius

x1² + y1² - r² = 9

(-3)² + 1² - r² = 9

9 + 1 - r² = 9

1 - r² = 0

r² = 1

r = 1

so, the standard circle equation is

(x + 3)² + (y - 1)² = 1