Question

Please answer this(i don't need a graph but I'll appreciate if you do one for each question)

thank you! :)​

Answer 1

Exercise 2:
a) Opens upward
b) h = 2, k = 0
c) x = 2
d) Domain: (-∞, ∞), Range: [0, ∞)

Exercise 3:
a) Opens upward
b) h = 1, k = -4
c) x = 1
d) Domain: (-∞, ∞), Range: [-4, ∞)

Explanation:

Exercise 2

Given quadratic equation:

`y=2(x-2)^2`

The parabola opens upward as the leading coefficient is positive.

The equation is already in vertex form, y = a(x - h)² + k, with the vertex (h, k) being (2, 0). However, as the question necessitates the use of specific formulas to determine the values of h and k, we should first expand the given equation into standard form, y = ax² + bx + c:

`y=2(x-2)^2`

`y=2(x-2)(x-2)`

`y=2(x^2-4x+4)`

`y=2x^2-8x+8`

Therefore:

• a = 2
• b = -8
• c = 8

To determine the vertex (h, k) using the provided formulas, plug in the values of a, b, and c into the equations:

`h=(-b)/(2a)=(-(-8))/(2(2))=(8)/(4)=2`

`k=(4ac-b^2)/(4a)=(4(2)(8)-(-8)^2)/(4(2))=(64-64)/(8)=(0)/(8)=0`

Therefore, the vertex is (2, 0).

The equation of the axis of symmetry for a vertical parabola is x = h, where h represents the x-coordinate of the vertex. Therefore, in this case, the equation of the axis of symmetry is x = 2.

The domain of a function is the set of all possible input values (x-values) for which the function is defined. The domain of a quadratic function is all real values of x, so the domain of y = 2(x - 2)² is (-∞, ∞).

The range of a function is the set of all possible output values (y-values) for which the function is defined. Given that the minimum value of y is the y-coordinate of the vertex, the range is restricted [0, ∞).

`\hrulefill`

Exercise 3

Given quadratic equation:

`y=x^2-2x-3`

The parabola opens upward as the leading coefficient is positive.

The given equation is in standard form, y = ax² + bx + c. Therefore, the values of a, b and c are:

• a = 1
• b = -2
• c = -3

To determine the vertex (h, k) using the provided formulas, plug in the values of a, b, and c into the equations:

`h=(-b)/(2a)=(-(-2))/(2(1))=(2)/(2)=1`

`k=(4ac-b^2)/(4a)=(4(1)(-3)-(-2)^2)/(4(1))=(-12-4)/(4)=(-16)/(4)=-4`

Therefore, the vertex is (1, -4).

The equation of the axis of symmetry for a vertical parabola is x = h, where h represents the x-coordinate of the vertex. Therefore, in this case, the equation of the axis of symmetry is x = 1.

The domain of a function is the set of all possible input values (x-values) for which the function is defined. The domain of a quadratic function is all real values of x, so the domain of y = x² - 2x - 3 is (-∞, ∞).

The range of a function is the set of all possible output values (y-values) for which the function is defined. Given that the minimum value of y is the y-coordinate of the vertex, the range is restricted [-4, ∞).