# STANDARD FORM OF A QUADRATIC FUNCTION

The standard form of a quadratic function is

y  =  ax2 + bx + c

where a, b and c are real numbers, and a  ≠  0.

## Using Vertex Form to Derive Standard Form

Write the vertex form of a quadratic function.

y  =  a(x - h)2 + k

Square the binomial.

y  =  a(x2 - 2xh + h2) + k

y  =  ax2 - 2ahx + ah2 + k

The equation y  =  ax2 - 2axh + ah2 + k is a quadratic function in standard form with

a  =  a

b  =  - 2ah

c  =  ah2 + k

The vertex of a quadratic function is (h, k), so to determine the x-coordinate of the vertex, solve b = -2ah for h.

b  =  -2ah

Divide each side by -2a.

- b / 2a  =  h

Because h is the x-coordinate of the vertex, we can use this value to find the y-value, k, of the vertex.

Example :

Find the vertex of the quadratic function :

f(x)  =  x2 - 8x + 12

Solution :

Step 1 :

Identify the coefficients a, b and c.

Comparing

ax2 + bx + c  and  x2 - 8x + 12,

we get

a  =  1, b  =  -8  and  c  =  12

Step 2 :

Solve for h, the x-coordinate of the vertex.

h  =  - b / 2a

Substitute.

h  =  - (-8) / 2(1)

h  =  8 / 2

h  =  4

Step 3 :

Substitute the value of h into the equation for x to find k, the y-coordinate of the vertex.

f(4)  =  42 - 8(4) + 12

f(4)  =  16 - 32 + 12

f(4)  =  - 4

So, the vertex of the given quadratic function is

(h, k)  =  (4, -4)

## Graphing a Quadratic Function in Standard Form

Example :

f(x)  =  x2 - 4x + 8

Solution :

Step 1 :

Identify the coefficients a, b and c.

Comparing

ax2 + bx + c  and  x2 - 4x + 8,

we get

a  =  1, b  =  -4  and  c  =  8

Step 2 :

Find the vertex of the quadratic function.

The x-coordinate of the vertex can be determined by

h  =  - b / 2a

Substitute.

h  =  - (-4) / 2(1)

h  =  4 / 2

h  =  2

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

f(2)  =  22 - 4(2) + 8

f(2)  =  4 - 8 + 8

f(2)  =  4

So, the vertex is

(h, k)  =  (2, 4)

Step 3 :

Find the axis of symmetry of the quadratic function.

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

In the vertex (2, 4), the x-coordinate is 2.

So, the axis of symmetry is

x  =  2

Step 4 :

Find the y-intercept of the quadratic function.

To find the y-intercept, put x = 0.

f(0)  =  02 - 4(0) + 8

f(0)  =  8

Step 5 :

Find a point symmetric to the y-intercept across the axis of symmetry.

Because (0, 8) is point on the parabola 2 units to the left of the axis of symmetry, x  =  2, (4, 8) will be a point on the parabola 2 units to the right of the axis of symmetry.

Step 6 :

Sketch the graph.

Once we have three points associated with the quadratic function, we can sketch the parabola based on our knowledge of its general shape.

## Interpreting the Graph of a Quadratic Function

Example :

The graph of a quadratic function

f(x)  =  - 10x2 + 700x - 6000

shows the profit, a company earns for selling items at different prices. Find the maximum profit that the company can expect to earn.

Solution :

The x-axis shows the selling price and the y-axis shows the profit.

The maximum y-value of the profit function occurs at the vertex of its parabola. Find the vertex of the parabola.

Use the function to find the x-coordinate and y-coordinate of the vertex.

Find the x-coordinate of the vertex.

h  =  - b / 2a

Substitute.

h  =  - 700 / 2(-10)

h  =  - 700 / (-20)

h  =  35

Find the y-coordinate of the vertex.

y  =  - 10x2 + 700x - 6000

Substitute  x  =  35.

y  =  - 10(35)2 + 700(35) - 6000

Simplify.

y  =  - 12250 + 24500 - 6000

y  =  6250

So, the vertex is

(h, k)  =  (35, 6250)

So, the selling price of \$35 per item gives the maximum profit of \$6,250.

## Writing the Equation of a Parabola Given Three Points

Example :

Find the equation of a parabola that passes through the points :

(-2, 0), (3, -10) and (5, 0)

Solution :

Step 1 :

Write the three equations by substituting the given x and y-values into the standard form of a parabola equation,

y  =  ax2 + bx + c

Substitute (-2, 0).

0  =  a(-2)2 + b(-2) + c

0  =  4a - 2b + c

Substitute (3, -10).

-10  =  a(3)2 + b(3) + c

-10  =  9a + 3b + c

Substitute (5, 0).

0  =  a(5)2 + b(5) + c

0  =  25a + 5b + c

Step 2 :

Solve the system :

0  =  4a - 2b + c

-10  =  9a + 3b + c

0  =  25a + 5b + c

Solving the above system using elimination method,  we will get

a  =  1,  b  =  - 3  and  c  =  - 10

Step 3 :

Substitute 1 for a, -3 for b, and -10 for c in the standard form of quadratic equation.

Hence, the equation of a parabola is

y  =  x2 - 3x - 10

Step 4 :

Confirm that the graph of the equation passes through the given three points.

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