Question

write an equation for the parabola whose vertex is at (2,6) and which passes through (4,-1)

Answer 1

Answer:

Step-by-step explanation:

To move the vertex up to (2,6), we write y = (x - 2)^2 + 6

y = x^2 - 4x + 4 + 6

y = x^2 - 4x + 10

When x = 4,

y = 4^2 - 4 * 4 + 10 = 16 - 16 + 10 = 10

We want that to be -4, so we are going to divide

-4 = 10 * p

p = -4/10 = -2/5

y = (-2/5) * (x^2 - 4x + 10)

y = (-2/5)x^2 + (8/5)x - 4

Answer 2

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4 answers · Mathematics 

 Best Answer

We know that the standard parabola is y = x^2. This has a vertex of (0,0) 

To move the vertex up to (2,6), we write y = (x - 2)^2 + 6 

y = x^2 - 4x + 4 + 6 
y = x^2 - 4x + 10 

When x = 4, 
y = 4^2 - 4 * 4 + 10 = 16 - 16 + 10 = 10 
We want that to be -4, so we are going to divide 
-4 = 10 * p 
p = -4/10 = -2/5 

y = (-2/5) * (x^2 - 4x + 10) 
y = (-2/5)x^2 + (8/5)x - 4