SOLUTION: Find the vertex of the graph of the equation x - y^2 + 8y = -4y^2 + 15y + 16.
Algebra.Com
Question 1209270: Find the vertex of the graph of the equation x - y^2 + 8y = -4y^2 + 15y + 16.
Answer by asinus(45) (Show Source): You can put this solution on YOUR website!
**1. Rewrite the Equation**
* Start by rearranging the equation to isolate 'x':
x = y² - 8y + 4y² - 15y + 16
x = 3y² - 23y + 16
**2. Complete the Square**
* **Factor out the coefficient of y²:**
x = 3(y² - (23/3)y) + 16
* **Inside the parentheses, add and subtract the square of half the coefficient of y:**
x = 3(y² - (23/3)y + (23/6)² - (23/6)²) + 16
* **Rewrite as a perfect square trinomial:**
x = 3[(y - 23/6)² - 529/36] + 16
* **Distribute the 3:**
x = 3(y - 23/6)² - 529/12 + 16
* **Simplify:**
x = 3(y - 23/6)² - 145/12
**3. Identify the Vertex**
* The equation is now in vertex form: x = a(y - k)² + h
* Where (h, k) represents the vertex of the parabola.
* In this case:
* h = -145/12
* k = 23/6
**Therefore, the vertex of the graph is (-145/12, 23/6).**
RELATED QUESTIONS
If RS=8y+4, ST=4y+8, and RT=15y-9, Find the value of... (answered by jim_thompson5910)
Find the vertex, focus, and directrix of the parabola given by the equation (x-1)^2 =... (answered by lwsshak3)
Find the vertex, focus, and directrix of the parabola X^2-4X- 4Y +16=0. Solve and graph (answered by josgarithmetic)
Write the equation of the circle in standard form find the center, radius, intercepts and (answered by lwsshak3)
name the vertex of the parabola with the equation y^2 - 8y + 18 =... (answered by lwsshak3)
find the vertex, focus, and directrix. Then draw the graph
1), {{{x^2+6x=8y-1}}}
2), (answered by josgarithmetic)
GRAPH THE EQUATION y = 1/2x + 3/2
2) y = x + 2
x + 2y = 16 x= __________... (answered by drj)
Find the vertex, focus, and directrix of the parabola given by the equation x^2 + 4x + 4y (answered by KMST)
Find the center, vertices, foci, and asymtotes of the hyperbola
4x^2-y^2-8x+2Y-1=0 AND... (answered by Edwin McCravy)