# DOMAIN RANGE VERTEX AND ZEROES OF QUADRATIC FUNCTION

Domain :

Domain of a quadratic function is always (-∞, ∞) because quadratic function will always extent forever in either direction along the x-axis.

Range :

Range is set of all possible outputs.

Vertex :

The vertex is the turning point of the parabola.

If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function.

If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value.

Axis of symmetry :

The axis of symmetry is the vertical line, which divides the parabola into two equal parts.

Zeroes :

To find zeroes of the polynomial, we have to replace y by 0.

A real number k of a quadratic polynomial p(x) is 0 if

p(k) = 0

Find the important information and sketch.

(i) Vertex form

(ii)  Max/Min

(iii) Axis of symmetry

(iv) Domain

(v) Range

(vi) Zeroes

(vii) y-intercept

Problem 1 :

f(x) = x2 + 4x + 8

Solution :

Let f(x) = y

y = x2 + 4x + 8

(i) Vertex form :

Converting the equation into vertex for

y = a(x - h)2 + k

Writing the coefficient of x as multiple of 2.

y = x2 + 2 ⋅ x ⋅ 2 + 22 - 22 + 8

y = (x + 2)2 - 22 + 8

y = (x + 2)2 - 4 + 8

y = (x + 2)2 + 4

(h, k) ==> (-2, 4)

(ii) Maximum / Minimum :

y = 1(x + 2)2 + 4

Here a = 1 > 0, so the parabola opens up.

The given quadratic function will have minimum value.

(iii)  Axis of symmetry :

x = -b/2a

f(x) = x2 + 4x + 8

Comparing the given equation with y = ax2 + bx + c

a = 1, b = 4 and c = 8

x = -4/2(1)

x = -2

(iv) Domain :

Domain of the function is -∞ < x <

(v) Range :

Since the minimum value is 4, the curve will go below 4. So the range is y    4.

(vi) Zeroes :

x2 + 4x + 8 = 0

Since this quadratic function is not factorable, we will check the nature of roots.

Nature of roots = b2 - 4ac

= 42 - 4(1)(8)

= 16 - 32

= -16 < 0

So, the quadratic function has no real roots.

(vii) y-intercept :

y = x2 + 4x + 8

To find y-intercept, put x = 0

y = 02 + 4(0) + 8

y = 8

So, the y-intercept is (0, 8).

Problem 2 :

f(x) = -4(x - 2)2 + 4

Solution :

(i) Vertex form :

The given questions is in vertex form.

y = -4(x - 2)2 + 4

(h, k) ==> (2, 4)

(ii) Maximum / Minimum :

y = -4(x - 2)2 + 4

Here a = -4< 0, so the parabola opens down.

The given quadratic function will have maximum value.

(iii)  Axis of symmetry :

Axis of symmetry will be at vertex.

x = -2

(iv) Domain :

Domain of the function is -∞ < x <

(v) Range :

Since the minimum value is 4, the curve will go below 4. So the range is y    4.

(vi) Zeroes :

-4(x - 2)2 + 4 = 0

-4(x - 2)2 = -4

(x - 2)2 = 1

Taking square roots on both sides, we get

x - 2 = √1

x - 2 = ±1

x - 2 = 1 and x - 2 = -1

x = 1 + 2 and x = -1 + 2

x = 3 and x = 1

So, zeroes are 1 and 3.

(vii) y-intercept :

y =  -4(x - 2)2 + 4

To find y-intercept, put x = 0

y =  -4(0 - 2)2 + 4

y = -4(4) + 4

y = -16 + 4

y = -12

So, the y-intercept is (0, -12).

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