VERTEX AXIS OF SYMMETRY MAX MIN OF QUADRATIC FUNCTION WORKSHEET

How to find vertex, axis of symmetry, maximum or minimum of quadratic function ?

Vertex of the parabola :

The vertex of a parabola is a point at which the parabola makes its sharpest turn. Vertex of the parabola can be found in two ways.

i) Converting the given quadratic function into vertex form.

ii) Using the formula.

Axis of symmetry :

The axis of symmetry is the vertical line that goes through the vertex of a parabola so the left and right sides of the parabola are symmetric.

Maximum or minimum :

Maximum or minimum will be at vertex.

If the parabola opens up, then it will have minimum.

If the parabola opens down, then we will have maximum.

For the questions given below, find the following.

a = ___ b = ___ c = ___

i)  Y-intercept _____

ii)  Vertex (___ , ___)

iii)  Equation of Axis of Symmetry: X = _____

iv)  Opens Up or Opens Down

v)  Maximum or Minimum

vi)  Max/Min Value _____

qua-fun-q1

Problem 1 :

y = -2x2 + 1

Solution:

y = -2x2 + 1

Here a = -2, b = 0 and c = 1

y-intercept:

To find the y-intercept put x = 0.

y = -2(0)2 + 1

y = 1

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=-02(-2)x=-0-4x=0

Substitute the value of x into the equation to find the y-coordinate of the vertex, k:

y = -2(0)2 + 1

y = 1

So, the vertex is (h, k) = (0, 1).

Axis of symmetry:

Axis of symmetry of a quadratic function can be determined by the x-coordinate of the vertex.

So, the axis of symmetry is x = 0.

Opens up/Opens down:

a < 0, the parabola opens down.

Max/Min value:

The maximum value is 1.

qua-fun-s1.png

Problem 2 :

y = x2 + 2x + 1

Solution:

Finding a, b and c:

a = 1, b = 2 and c = 1

y-intercept:

To find the y-intercept put x = 0.

y = (0)2 + 2(0) + 1

y = 1

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=-22(1)x=-1

y = (-1)2 + 2(-1) + 1

y = 1 - 2 + 1

y = 0

So, the vertex is (h, k) = (-1, 0).

Axis of symmetry:

Axis of symmetry is x = -1.

Opens up/Opens down:

a > 0, the parabola opens up.

Max/Min value:

The minimum value is 0.

vertex-max-minq2

Problem 3 :

y=-14x2-1

Solution:

Finding a, b and c :

a = -1/4, b = 0 and c = -1

y-intercept:

To find the y-intercept put x = 0.

y = -1/4(0)2 - 1

y = -1

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=-02(-1/4)x=0

When x = 0

y = -1/4(0)2 - 1

y = 0 - 1

y = -1

So, the vertex is (h, k) = (0, -1).

Axis of symmetry:

Axis of symmetry is x = 0.

Opens up/Opens down:

a < 0, the parabola opens down.

Max/Min value:

The maximum value is -1.

qua-fun-a3

Problem 4 :

y=-12x2-4x-1

Solution:

Finding a, b and c:

a = -1/2, b = -4 and c = -1

y-intercept:

To find the y-intercept put x = 0.

y = -1/2(0)2 - 4(0) - 1

y = -1

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=--42(-1/2)x=-4

When x = -4

y = -1/2(-4)2 - 4(-4) - 1

y = -8 + 16 - 1

y = 7

So, the vertex is (h, k) = (-4, 7).

Axis of symmetry:

Axis of symmetry is x = -4.

Opens up/Opens down:

a < 0, the parabola opens down.

Max/Min value:

The maximum value is 7.

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Problem 5 :

y = 3x2 - 12x - 5

Solution:

Finding a, b and c:

a = 3, b = -12 and c = -5

y-intercept:

To find the y-intercept put x = 0.

y = 3(0)2 - 12(0) - 5

y = -5

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=--122(3)x=2

When x = 2, 

y = 3(2)2 - 12(2) - 5

y = 12 - 24 - 5

y = -17

So, the vertex is (h, k) = (2, -17).

Axis of symmetry:

Axis of symmetry is x = 2.

Opens up/Opens down:

a > 0, the parabola opens up.

Max/Min value:

The minimum value is -17.

qua-fun-a5

Problem 6 :

y = -x2 - 8x - 13

Solution:

Finding a, b and c:

a = -1, b = -8 and c = -13

y-intercept:

To find the y-intercept put x = 0.

y = -(0)2 - 8(0) - 13

y = -13

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=--82(-1)x=-4

When x = -4

y = -(-4)2 - 8(-4) - 13

y = -16 + 32 - 13

y = 3

So, the vertex is (h, k) = (-4, 3).

Axis of symmetry:

Axis of symmetry is x = -4.

Opens up/Opens down:

a < 0, the parabola opens down.

Max/Min value:

The maximum value is 3.

qua-fun-a6

Problem 7 :

y = x2 + 2x - 2

Solution:

Finding a, b and c:

a = 1, b = 2 and c = -2

y-intercept:

To find the y-intercept put x = 0.

y = (0)2 + 2(0) - 2

y = -2

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=-22(1)x=-1

When x = -1

y = (-1)2 + 2(-1) - 2

y = 1 - 2 - 2

y = -3

So, the vertex is (h, k) = (-1, -3).

Axis of symmetry:

Axis of symmetry is x = -1.

Opens up/Opens down:

a > 0, the parabola opens up.

Max/Min value:

The minimum value is -3.

qua-fun-a7

Problem 8 :

y = 2x2 - 12x + 14

Solution:

Finding a, b and c:

a = 2, b = -12 and c = 14

y-intercept:

To find the y-intercept put x = 0.

y = 2(0)2 - 12(0) + 14

y = 14

Vertex:

The x-coordinate of the vertex can be determined by

x=-b2ax=--122(2)x=3

When x = 3

y = 2(3)2 - 12(3) + 14

y = 18 - 36 + 14

y = -4

So, the vertex is (h, k) = (3, -4).

Axis of symmetry:

Axis of symmetry is x = 3.

Opens up/Opens down:

a > 0, the parabola opens up.

Max/Min value:

The minimum value is -4.

qua-fun-a8

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