We have a quadratic function written in standard form.

y = a x^{2} + b x + c

This kind of equation can give us a lot of information about the parabola by observing the values of

a,

b,

and

c .

y = - x^{2} + 2 x + 5 ⇕ y = - 1 x^{2} + (2) x + 5

We see that for the given equation,

a = - 1,

b = 2,

and

c = 5 .

$x -$ value of the Vertex

Consider the point at which the curve of the parabola changes direction.

This point is the vertex of the parabola, and defines the axis of symmetry. If we want to calculate the

x -

value of this point, we can substitute the given values of

a

and

b

into the expression

- \frac{b}{2 a}

and simplify.

- \frac{b}{2 a}

SubstituteII

$a = - 1$ , $b = 2$

- \frac{2}{2 ( - 1 )}

▼

Simplify

MultPosNeg

$a (- b) = - a \cdot b$

- \frac{2}{- 2}

MoveNegFracToNum

Put minus sign in numerator

\frac{- 2}{- 2}

QuotOne

$\frac{a}{a} = 1$

1

Remember, the axis of symmetry is the vertical line that passes through the vertex, dividing a parabola into two mirror images. Since every point on this line will have the same

x -

coordinate as the vertex, we can form its equation.

x = 1

$y -$ value of the Vertex

The point at which the graph of a parabola changes direction also defines the maximum or minimum point of the graph. Whether the parabola has a minimum or maximum is determined by the value of $a .$

Since the given value of

a

is negative, the parabola has a maximum value at the vertex. To find this value, think of

y

as a function of

x,

y = f (x) .

By substituting the

x -

value of the vertex into the given equation and simplifying, we will get the

y -

value of the vertex.

y = - x^{2} + 2 x + 5

Substitute

$x = 1$

y = - (1)^{2} + 2 (1) + 5

▼

Simplify right-hand side

CalcPow

Calculate power

y = - 1 + 2 (1) + 5

MultByOne

$a \cdot 1 = a$

y = - 1 + 2 + 5

AddTerms

Add terms

y = 6

The Vertex

Given the standard form of a parabola, the

(x, y)

coordinates of its vertex can be expressed in terms of

a

and

b .

(x, y) \Leftrightarrow (- \frac{b}{2 a}, f (- \frac{b}{2 a}))

We have already calculated both of these values above, so we know that the vertex lies on the point

(1, 6) .

Range

Recall that we already found the parabola to open downwards from a maximum point. The range of the function is therefore all numbers less than or equal to the maximum value.

Range : y \leq 6

Extra

A Common Mistake

One common mistake when identifying the key features of a parabola algebraically is forgetting to include the negatives in the values of these constants. The standard form is addition only, so any subtraction must be treated as negative values of

a,

b,

or

c .

Let's look at an example.

\underline{a x^{2} + b x + c} y = 3 x^{2} - 4 x - 2 ⇕ y = 3 x^{2} + (- 4 x) + (- 2)

In this case, the values of

a,

b,

and

c

are

3,

- 4,

and

- 2 .

They are not

3,

4,

and

2 .

a = 3, b = 4, c = 2 \times a = 3, b = - 4, c = - 2 ✓

x-value of the Vertex

y-value of the Vertex

The Vertex

Range

Extra

$x -$ value of the Vertex

$y -$ value of the Vertex

Exercise