y = ax^2 + bx + 5; (-1,4), Finding a and b?

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y(-1) = a - b + 5 = 4
a - b = -1 => a = b - 1
-b/a = -b/(b - 1)
c/a = 5
a = 1
b = 2
y = x^2 + 2x + 5...Show more

If (-1,4) is the vertex then the function can be written as y = (x + p)^2 + q

The x co-ordinate makes the bracket zero and q is the y co-ordinate of the vertex.

So, y = (x +1)^2 + 4

Expanding and simplifying gives:

y = x^2 + 2x + 5, so a = 1 and b = 2 ....resulting in a parabola with minimum point at (-1,4)...Show more

If (-1,4) is onto your parabola then
4 = a*(-1)^2 - b*(-1) + 5
a-b = -1
b = a+1
Then
y = ax^2 +(a+1)x+5
This is a parabolas family and passes to (-1,4)
You must have 2 points to find a and b for parabola y = ax^2 + bx + 5...Show more

y = ax^2 + bx + 5; (-1,4)
Find a and b

4 = a - b + 5

a -b = 4 -5 .................... [1]

Analysing [1], we get,

a = 4, and b = -5 >=================< ANSWER...Show more

y = ax^2 + bx + 5; (-1,4)

y = (x + 1)(x - 4)
y = x^2 - 3x - 4
Find a and b

???...Show more