What’s is part A?? .....

Answer:
A. 3
B. 4
C. 1
D. 2
Step-by-step explanation:
Consider all equations:
A. [tex]y=x^2 -6x+8[/tex]
This is the equation of parabola with vertex at point
[tex]x_v=\dfrac{-b}{2a}=\dfrac{-(-6)}{2\cdot 1}=3\\ \\y_v=3^2-6\cdot 3+8=9-18+8=-1[/tex]
The y-intercept is at point
[tex]x=0\\ \\y=0^2-6\cdot 0+8=8[/tex]
Since
[tex]x^2 -6x+8=x^2-4x-2x+8=x(x-4)-2(x-4)=(x-2)(x-4)[/tex],
the x-intercepts are at points (2,0) and (4,0)
The leading coefficient is 1 > 0, then the parabola opens upwards.
Hence, the graph of this parabola is 3.
B. [tex]y=(x-6)(x+8)[/tex]
This parabola has two x-intercepts at points (6,0) and (-8,0).
The leading coefficient is 1 > 0, then the parabola opens upwards.
The only possible choice is parabola 4.
C. [tex]y=(x-6)^2+8[/tex]
This parabola has the vertex at point (6,8), opens upwards, therefore does not intersect the x-axis.
The graph of this parabola is 1.
D. [tex]y=-(x+8)(x-6)[/tex]
This parabola has two x-intercepts at points (6,0) and (-8,0).
The leading coefficient is -1 > 0, then the parabola opens downwards.
The only possible choice is parabola 2.
Answer:
Equation A matches Graph 3, because the parabola is opened upward and intersected the x-axis at points (2 , 0) and (4 , 0)
Equation B matches Graph 4, because the parabola is opened upward, intersected the x-axis at point (-8 , 0) , (6 , 0) and the y-axis at point (0 , -48)
Equation C matches Graph 1, because the parabola is above the x-axis
Equation D matches Graph 2, because the parabola is opened downward
Step-by-step explanation:
The quadratic function f(x) = ax² + bx + c is represented graphically by a parabola which has a vertex point (h , k)
The vertex form of the quadratic function is f(x) = (x - h)² + k
Equation A
∵ y = x² - 6x + 8
∵ a = 1
- By using the 1st note above
∴ The graph of the equation is a parabola opened upward
- Lets find its zeros (Put y = 0)
∵ x² - 6x + 8 = 0
- Factorize it into two factors
∴ (x - 2)(x - 4) = 0
- Equate each factor by 0
∵ x - 2 = 0 ⇒ add 2 to both sides
∴ x = 2
∵ x - 4 = 0 ⇒ add 4 to both sides
∴ x = 4
∴ Its zeroes are 2 and 4
- That means the parabola intersects the x-axis at 2 and 4
∴ The graph of equation A is figure 3
Equation A matches Graph 3, because the parabola is opened upward and intersected the x-axis at points (2 , 0) and (4 , 0)
Equation B
∵ y = (x - 6)(x + 8)
- Equate each factor by 0
∵ x - 6 = 0 ⇒ add 6 to both sides
∴ x = 6
∵ x + 8 = 0 ⇒ subtract 8 from both sides
∴ x = -8
∴ Its zeroes are 2 and 4
- That means the parabola intersects the x-axis at points (-8 , 0)
and (6 , 0)
∵ (x - 6)(x + 8) = x² + 8x - 6x - 48
- Add like terms
∴ (x - 6)(x + 8) = x² + 2x - 48
∴ y = x² + 2x - 48
- Find the value of a and c
∴ a = 1 and c = -48
∵ a is positive
∴ The parabola is opened upward
∵ c = -48
∴ The parabola intersects the y-axis at point (0 , -48)
∴ The graph of equation B is figure 4
Equation B matches Graph 4, because the parabola is opened upward intersected the x-axis at point (-8 , 0) , (6 , 0) and the y-axis at point (0 , -48)
Equation C
∵ y = (x - 6)² + 8
- The equation is in the vertex form
∴ The coordinates of the vertex point are h = 6 and k = 8
∴ The vertex of the parabola is (6 , 8)
- By using the 4th note above
∵ k > 0
∴ The parabola does not intersect the x-axis
∴ the parabola is above the x-axis
∴ The graph of equation C is figure 1
Equation C matches Graph 1, because the parabola is above the x-axis
Equation D
∵ y = -(x + 8)(x - 6)
(x + 8)(x - 6) = x² - 6x + 8x - 48
- Add like terms
∴ -(x - 6)(x + 8) = -(x² + 2x - 48)
- Multiply each term by (-)
∴ -(x - 6)(x + 8) = - x² - 2x + 48
∴ y = -x² - 2x + 48
- Find the value of a
∴ a = -1
- Use the 1st rule above
∵ a < 0
∴ The parabola is opened downward
∴ The graph of equation D is figure 2
Equation D matches Graph 2, because the parabola is opened downward