For the quadratic function $f (x) = a x^{2} + b x + c,$ the $y -$ coordinate of the vertex is the maximum value of the function when $a < 0 .$

Let's identify the values of

a,

b,

and

c

in the given quadratic function.

f (x) = - 2 x^{2} + 4 x - 1 ⇕ f (x) = - 2 x^{2} + 4 x + (- 1)

We can see above that $a = - 2,$ $b = 4,$ and $c = - 1 .$ We will now use these values to find the desired information.

Since

a = - 2

is less than

0,

the parabola will open downwards. This means it will have a maximum value, which is given by

f (- \frac{b}{2 a}) .

Before we find the value of the function at this point, we need to substitute

a = - 2

and

b = 4

in

- \frac{b}{2 a} .

- \frac{b}{2 a}

▼

Substitute values and evaluate

$a = - 2$ , $b = 4$

- \frac{4}{2 ( - 2 )}

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

- \frac{4}{- 4}

$- \frac{a}{- b} = \frac{a}{b}$

\frac{4}{4}

Calculate quotient

1

Now we have to calculate

f (1) .

To do so, we will substitute

1

for

x

in the given function.

f (x) = - 2 x^{2} + 4 x - 1

$x = 1$

f (1) = - 2 (1)^{2} + 4 (1) - 1

▼

Simplify right-hand side

Calculate power

f (1) = - 2 (1) + 4 (1) - 1

f (1) = - 2 + 4 - 1

Add and subtract terms

f (1) = 1

This tells us that the maximum value of the function is

1 .

Unless there are any specified restrictions on the

x -

values, the domain of a quadratic function is all real numbers. Therefore, the domain of this function is all real numbers. Furthermore, since

a = - 2

is less than

0,

the range is all values less than or equal to the maximum value,

1 .

Domain : Range : All real numbers y \leq 1

Since

a = - 2

is less than

0,

the function increases to the left of the maximum value and decreases to the right of the maximum value, which we know occurs at

1 .

Increasing Interval : Decreasing Interval : To the left of 1 To the right of 1

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