Find the vertex of the parabola. $x ^ { 2 } + 4 x - 8 y + 36 = 0$
A) $V ( 6 , - 1 )$
B) $V ( - 2,4 )$
C) $V ( 5 , - 2 )$
D) $V ( - 2 , - 6 )$
E) $V ( - 2,8 )$

Question

Quizplus · Accepted Answer

To find the vertex of the parabola given by $x^2 + 4x - 8y + 36 = 0$, we first need to express the equation in vertex form. We can start by isolating the y-term:$$8y = x^2 + 4x + 36$$Then, divide by 8:$$y = \frac{1}{8}x^2 + \frac{1}{2}x + 4.5$$To complete the square, we focus on the quadratic and linear terms in x:$$\frac{1}{8}x^2 + \frac{1}{2}x = \frac{1}{8}(x^2 + 4x)$$To complete the square, we add and subtract $(\frac{4}{2})^2 = 4$, inside the parenthesis, considering the coefficient of $x^2$ which is 1 after factoring out $\frac{1}{8}$:$$\frac{1}{8}(x^2 + 4x + 4 - 4) + 4.5$$This simplifies to:$$\frac{1}{8}(x + 2)^2 - \frac{1}{2} + 4.5$$Or:$$y = \frac{1}{8}(x + 2)^2 + 4$$Thus, the vertex of the parabola is at $(-2, 4)$.

Find the Vertex of the Parabola x2+4x−8y+36=0x ^ { 2 } + 4 x - 8 y + 36 = 0x2+4x−8y+36=0

Find the Vertex of the Parabola $x ^ { 2 } + 4 x - 8 y + 36 = 0$