A quadratic curve cuts the x-axis at 3 and -4.It cuts the y-axis at Z. The equation of the graph is y=ax^2+bx+c. Find ALL the posible values of a,b and c and the coordinates of Point Z.
Coolguy -- be careful! You're stepping out of bounds.
Thanks Ms Sue For Repremanding Coolguy
You're welcome, Ackron.
y = k(x+4)(x-3)
now plug in x=0 to find z.
z = -12k
k = -z/12
So,
y = -z/12 (x^2+x-12)
= -z/12 x^2 - z/12 x + z
so, for each z there is only one possible value for each of a,b,c
To find the equation of the quadratic curve, we need to determine the values of a, b, and c. We can do this by using the given information about the x-intercepts and y-intercept.
1. x-intercepts:
The quadratic curve cuts the x-axis at 3 and -4. This means that when x = 3 and x = -4, y = 0.
When x = 3, y = 0, we can substitute these values into the quadratic equation:
0 = a(3)^2 + b(3) + c
0 = 9a + 3b + c
When x = -4, y = 0, we can substitute these values into the quadratic equation:
0 = a(-4)^2 + b(-4) + c
0 = 16a - 4b + c
2. y-intercept:
The quadratic curve cuts the y-axis at point Z, which means that when x = 0, y = Z.
Substituting the values into the quadratic equation:
Z = a(0)^2 + b(0) + c
Z = c
Therefore, the value of c is equal to the y-intercept, Z.
Now we have three equations:
1. 0 = 9a + 3b + c
2. 0 = 16a - 4b + c
3. c = Z
We can rearrange the first two equations to eliminate c:
a = (4b - Z)/16
a = (-3b + Z)/9
Solving these two equations simultaneously will give us the possible values of a and b:
(4b - Z)/16 = (-3b + Z)/9
Cross-multiplying:
9(4b - Z) = 16(-3b + Z)
36b - 9Z = -48b + 16Z
Bringing the variables to one side and the constants to the other:
36b + 48b = 16Z + 9Z
84b = 25Z
Simplifying:
b = (25Z)/84
b = (25Z)/84
Using the value of b, we can substitute it back into one of the original equations to get the value of a:
a = (-3((25Z)/84) + Z)/9
Simplifying:
a = (-75Z + 84Z)/252
a = 9Z/252
a = Z/28
Therefore, the possible values of a, b, and c are:
a = Z/28
b = (25Z)/84
c = Z
To find the coordinates of point Z, we need to substitute x = 0 into the equation of the curve:
y = a(0)^2 + b(0) + c
y = c
Therefore, the coordinates of point Z are (0, c), or specifically (0, Z).