PROPERTIES OF QUADRATIC EQUATIONS

Convert each quadratic function to its vertex form. Then, state the coordinates of the vertex

Problem 1 :

x² - 4 x - 5

Solution :

= x² - 4 x - 5

To convert the quadratic function from standard form to vertex form, we have to use the method of completing the square.

= x² - 2(x) (2) + 2² - 2² - 5

= (x - 2)² - 4 - 5

= (x - 2)² - 9

Comparing with y = a(x - h)² + k

Vertex (h, k) ==> (2, -9)

Problem 2 :

x² + x - 3

Solution :

= x² + x - 3

To convert the quadratic function from standard form to vertex form, we have to use the method of completing the square.

= x² + 2(x) (1/2) + (1/2)² - (1/2)² - 3

= (x + (1/2))² - (1/4) - 3

= (x + (1/2))² - 13/4

Comparing with y = a(x - h)² + k

Vertex (h, k) ==> (-1/2, -13/4)

Use the vertex formula, to find the coordinates of the vertex of each parabola.

Problem 3 :

x² + 6x + 3

Solution :

= x² + 6x + 3

Comparing with ax² + bx + c, a = 1, b = 6 and c = 3

x-coordinate of vertex :

x = -b/2a

x = -6/2(1)

x = -6/2

x = -3

y-coordinate of vertex :

f(-3) = (-3)² + 6(-3) + 3

= 9 - 18 + 3

= 12 - 18

= -6

So, the vertex is at (-3, -6).

Problem 4 :

(1/2)x² - 4x - 7

Solution :

= (1/2)x² - 4x - 7

Comparing with ax² + bx + c, a = 1/2, b = -4 and c = -7

x-coordinate of vertex :

x = -b/2a

x = -4/2(1/2)

x = -4

y-coordinate of vertex :

f(-4) = (1/2)(-4)² - 4(-4) - 7

= (1/2) (16) + 16 - 7

= 8 + 16 - 7

= 24 - 7

= 17

So, the vertex is at (-4, 17).

For each parabola, state its vertex, opening and shape. Then graph it and state the domain and range.

Problem 5 :

𝑓(𝑥) = 2𝑥² + 12𝑥 + 18

Solution :

𝑓(𝑥) = 2𝑥² + 12𝑥 + 18

𝑓(𝑥) = 2[𝑥² + 6𝑥] + 18

𝑓(𝑥) = 2[𝑥² + 2𝑥(3) + 3² - 3²] + 18

= 2[𝑥² + 2𝑥(3) + 3² - 9] + 18

= 2[(𝑥 + 3)² - 9] + 18

= 2(𝑥 + 3)² - 18 + 18

f(x) = 2(𝑥 + 3)²

Comparing with y = a(x - h)² + k

Vertex (h, k) ==> (-3, 0)

Direction of opening :

The parabola opens up and will have minimum value.

Domain and range :

Domain is -∞ ≤ x ≤ ∞

Range is 0 ≤ y  ≤ ∞

Problem 6 :

𝑓(𝑥) = -2𝑥² + 3𝑥 - 1

Solution :

𝑓(𝑥) = -2𝑥² + 3𝑥 - 1

𝑓(𝑥) = -2[𝑥² + (3/2)𝑥] - 1

𝑓(𝑥) = -2[𝑥² + 2x(3/4) + (3/4)² - (3/4)²] - 1

= -2[(𝑥 + (3/4))² - (3/4)²] - 1

= -2[(𝑥 + (3/4))² - (9/16)] - 1

= -2(𝑥 + (3/4))² + (9/8) - 1

= -2(𝑥 + (3/4))² + (1/8)

Comparing with y = a(x - h)² + k

Vertex (h, k) ==> (-3/4, 1/8)

Direction of opening :

The parabola opens down and will have maximum value.

Domain and range :

Domain is -∞ ≤ x ≤ ∞

Range is -∞ ≤ y  ≤ 1/8

For each quadratic function, state its zeros (roots), coordinates of the vertex, opening and shape. Then graph it and identify its extreme (minimum or maximum) value as well as where it occurs.

Problem 7 :

𝑓(𝑥) = (1/2)(x + 3)(x - 4)

Solution :

𝑓(𝑥) = (1/2)(x + 3)(x - 4)

Zeroes :

(1/2)(x + 3)(x - 4) = 0

x + 3 = 0 and x - 4 = 0

x = -3 and x = 4

Converting into vertex form :

𝑓(𝑥) = (1/2)(x² - x - 12)

= (1/2)(x² - 2 x (1/2) + (1/2)² - (1/2)²  - 12)

= (1/2)[x - (1/2))² - (1/4)] - 12

= (1/2)[x - (1/2))²] - (1/8) - 12

= (1/2)(x - (1/2))² - (97/8)

The parabola opens up, the minimum value is at

(1/2, -97/8)

True or false? Explain.

Problem 8 :

The domain and range of a quadratic function are both the set of real numbers.

Solution :

For every quadratic function domain is all real numbers

Problem 9 :

The graph of every quadratic function has exactly one 𝑦-intercept.

Problem 10 :

The graph of 𝑦 = −2(𝑥 −1)² −5 has no 𝑥𝑥-intercepts.

Problem 11 :

The maximum value of 𝑦𝑦 in the function 𝑦 = −4(𝑥−1)²+9 is 9. 

Problem 12 :

The value of the function 𝑓(𝑥)=𝑥²−2𝑥+1 is at its minimum when 𝑥=0.

48. The graph of 𝑓(𝑥)=9𝑥²+12𝑥+4 has one 𝑥-intercept and one 𝑦-intercept.

49. If a parabola opens down, it has two 𝑥-intercepts.

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