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1 answer:
Answer:
Center (1,3) and radius 6
Step-by-step explanation:
We must complete the square to find the center and radius of the circle.
First make sure the x and y squared terms have 1 as their coefficients. We also make sure x and y terms together.
We now create space between the x and y terms with parenthesis.
We complete the square by taking the middle terms -2x and the -6y - divide each and square them.
We add the squares to both sides.
Simplify.
And write the quadratics in factored form.
The center is (h,k) or (1,3). The radius is the square root of 36 which is 6.
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A graph of the equation shows the appropriate choice to be
C. 2
_____
If you would rather, you can look at the value of the discriminant. For the equation y = ax²+bx+c, the discriminant (d) is
d = b²-4ac
For your equation, this evaluates to
d = (-8)²-4(2)(5) = 64 -40 = 24
When the discriminant is positive, the function has two real roots (2 x-intercepts). When it is zero, there is only one x-intercept, and when it is negative, there are none (the roots are complex).
C. 2
_____
If you would rather, you can look at the value of the discriminant. For the equation y = ax²+bx+c, the discriminant (d) is
d = b²-4ac
For your equation, this evaluates to
d = (-8)²-4(2)(5) = 64 -40 = 24
When the discriminant is positive, the function has two real roots (2 x-intercepts). When it is zero, there is only one x-intercept, and when it is negative, there are none (the roots are complex).