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y = ax^2 + bx + 5; (-1,4), Finding a and b?
y = ax^2 + bx + 5; (-1,4)
Find a and b
(-1,4) is the vertex
7 Answers
- icemanLv 71 decade agoFavorite Answer
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
- Wayne DeguManLv 71 decade ago
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)
- Moise GunenLv 71 decade ago
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
- Fazaldin ALv 71 decade ago
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
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- RajendiranLv 61 decade ago
y = ax^2 + bx + 5; (-1,4)
y = (x + 1)(x - 4)
y = x^2 - 3x - 4
Find a and b
???
- margadonnaLv 44 years ago
Q2. The quadratic formula states 2 recommendations: they are (-b+ sqrt(b^2-4ac))/2a and (-b- sqrt(b^2-4ac))/2a. a stands for the coefficient of the squared term, b stands for the coefficinet of the 1st degree term, and c stands for the consistent term. subsequently as a effect a=2. b=a million, and c=-6 subsequently the recommendations are ( -a million+sqrt(a million+24))/4 = ( -a million +5)/4 = a million and (-a million -sqrt(a million+24))/4 = ( -a million -5)/4 = -3/2 Q1. here use the comparable formula (when you write the equation in the excellent suited sort, that's y^2+5y-14=0).
- Anonymous1 decade ago
replace the coords in the eqns:
4= a(-1)^2+b(-1) +5
4=a^2-b+5
a^2-b= -1
a^2 = b-1
a= ±√ (b-1)
b=a^2+1
Source(s): you should be having 2 coords or something to have the value of a and b it's just in an expression now.