a)Express y = 2 (x - 1) (x - 5) in the form y = ax^2 + bx + c.
b) From the above quadratic equation calculate the following:
I) x and y intercepts
II) Coordinate of the turning point
III) Is the graph a maximum or minimum turning point?
you can read the x-intercepts right off the original equation
once you expand it to the desired form, the y-intercept is just the value of c.
The vertex (turning point) is midway between the x-intercepts.
Since the coefficient of x^2 is positive, the parabola opens upward.
a) To express the quadratic equation y = 2(x - 1)(x - 5) in the form y = ax^2 + bx + c, we can follow these steps:
Step 1: Simplify the equation
Multiply the two binomials together:
y = 2(x^2 - 5x - x + 5)
Step 2: Combine like terms
y = 2(x^2 - 6x + 5)
Step 3: Expand further
y = 2x^2 - 12x + 10
Thus, the quadratic equation y = 2(x - 1)(x - 5) can be expressed in the form y = 2x^2 - 12x + 10.
b) Now let's calculate the following information from the given quadratic equation:
I) x and y intercepts:
To find the x-intercepts, set y = 0 and solve for x:
0 = 2x^2 - 12x + 10
You can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula here:
x = (-(-12) ± √((-12)^2 - 4(2)(10))) / (2(2))
x = (12 ± √(144 - 80)) / 4
x = (12 ± √64) / 4
x = (12 ± 8) / 4
So, we have two solutions for x:
x1 = (12 + 8) / 4 = 5
x2 = (12 - 8) / 4 = 1/2
Therefore, the x-intercepts are x = 5 and x = 1/2.
To find the y-intercept, set x = 0 and solve for y:
y = 2(0 - 1)(0 - 5)
y = 2(-1)(-5)
y = 10
Hence, the y-intercept is y = 10.
II) Coordinate of the turning point:
In general form, a quadratic equation can be written as y = ax^2 + bx + c. The turning point of a quadratic equation can be found using the formula:
x = -b / (2a)
Here, a = 2 and b = -12.
x = -(-12) / (2(2))
x = 12 / 4
x = 3
To find the y-coordinate of the turning point, substitute this value of x into the equation:
y = 2(3)^2 - 12(3) + 10
y = 2(9) - 36 + 10
y = 18 - 36 + 10
y = -8
Therefore, the coordinates of the turning point are (3, -8).
III) Is the graph a maximum or minimum turning point?
Since the coefficient of the x^2 term (a) is positive (a = 2 > 0), the graph opens upward, indicating a minimum turning point.