find the constants a, b and c such that or all values of x,
3x^2 - 5x + 1 = a(x+b)^2 +c
also whats the coordinate of the minimum point??
thanks for the help
Y = 3x^2 _ 5x + 1.
Vertex Form: y = a(x - h)^2 + k,
a = 3 = coefficient of x^2.
h = Xv = -b / 2a = 5/6 = 0.833.
Substitute 5/6 for x in the given Eq:
K = Yv=3*(5/6)^2 - 5*5/6 + 1 = -1.083.
Vertex Form: Y = 3(x - 5/6)^2 - 1.083.
V(h , k)=V(0.833 , -1.083) = Lowest
point.
=
To find the constants a, b, and c, we need to expand the right-hand side of the equation and compare it to the left-hand side.
Expanding the right-hand side:
a(x + b)^2 + c = a(x^2 + 2bx + b^2) + c = ax^2 + 2abx + ab^2 + c
Comparing the left-hand side to the expanded right-hand side:
3x^2 - 5x + 1 = ax^2 + 2abx + ab^2 + c
Equating the corresponding coefficients:
(ax^2) should be equal to (3x^2), which means a = 3.
(2abx) should be equal to (-5x), which means 2ab = -5.
(ab^2 + c) should be equal to 1, which means ab^2 + c = 1.
We have a system of equations:
a = 3
2ab = -5
ab^2 + c = 1
Using the first equation, we can substitute a = 3 into the second equation:
2(3)b = -5
6b = -5
b = -5/6
Substituting the value of b into the third equation:
(3)(-5/6)^2 + c = 1
25/12 + c = 1
c = 1 - 25/12
c = 12/12 - 25/12
c = -13/12
Therefore, the constants are:
a = 3, b = -5/6, c = -13/12
To find the coordinate of the minimum point, we need to consider the vertex of the quadratic function. The general form of a quadratic equation is ax^2 + bx + c, where the x-coordinate of the vertex is given by -b/(2a).
In this case, the quadratic equation is 3x^2 - 5x + 1. The x-coordinate of the vertex is -(-5)/(2*3) = 5/6.
Substituting this x-coordinate back into the original equation to find the y-coordinate:
y = 3(5/6)^2 - 5(5/6) + 1
y = 25/12 - 25/6 + 1
y = -13/12
Therefore, the coordinate of the minimum point is (5/6, -13/12).
To find the constants a, b, and c such that the given equation is true for all values of x, we need to expand the expression on the right side and then compare it to the left side of the equation.
Let's expand (x + b)^2:
(x + b)^2 = x^2 + 2bx + b^2
Now, let's multiply the right side of the equation by a:
a(x + b)^2 = a(x^2 + 2bx + b^2) = ax^2 + 2abx + ab^2
Now, let's compare this to the left side of the equation:
3x^2 - 5x + 1 = ax^2 + 2abx + ab^2 + c
By comparing the coefficients of x^2, x, and the constant term, we can find the values of a, b, and c.
Coefficients of x^2:
3 = a
Coefficients of x:
-5 = 2ab => -5/2 = ab => a = -5/2b
Constant term:
1 = ab^2 + c
Now we have a system of equations:
a = 3
a = -5/2b
1 = ab^2 + c
By substituting the value of a from the first equation into the second equation, we can solve for b:
-5/2b = 3 => -5/2 = 3b => b = -5/6
Substituting the values of a and b into the third equation, we can solve for c:
1 = (3)(-5/6)^2 + c => 1 = 25/12 + c => c = 1 - 25/12 => c = -7/12
Therefore, the constants are:
a = 3
b = -5/6
c = -7/12
To find the coordinates of the minimum point, we need to determine the vertex of the quadratic equation. Since the coefficient of x^2 is positive (3 in this case), the parabola opens upwards and the vertex represents the minimum point.
The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
Substituting the values of a and b, we get:
x = -(-5/6) / (2*3) => x = 5/36
To find the y-coordinate of the vertex, we substitute this x-value into the equation:
y = 3(5/36)^2 - 5(5/36) + 1 = -187/72
Therefore, the coordinates of the minimum point are:
(x, y) = (5/36, -187/72)