Question

Given the equation y=14x2−5, the minimum is ___ and the range is ___
and when x=-14, then f(x)= ____

Answer 1

Answer:

The minimum is -5

The range is $y \ge -5$

f(14) = 2739

Step-by-step explanation:

Given

$y = 14x^2 - 5$

Solving (a): The minimum

A quadratic function is represented as:

$y = ax^2 + bx + c$

If a > 0, then the function has a minimum

By comparison

$a = 14$ --- the function has a minimum

$b = 0$

$c = -5$

To calculate the minimum, we first calculate the following is calculated as:

$m = -\frac{b}{2a}$

So, we have:

$m = -\frac{0}{2*14}$

$m = 0$

So, the minimum is at f(m)

We have: $y = 14x^2 - 5$

$f(0) = 14 *0^2 - 5$

$f(0) = - 5$

Solving (b): The range

In (a), we have:

$f(0) = - 5$ --- the minimum

This implies that the smallest value of y on the graph is -5.

So, the range is:

$r = \{y|y\ge -5\}$

Solving (c): f(14)

We have:

$y = 14x^2 - 5$

So:

$f(14) = 14 * 14^2 - 5$

$f(14) = 2739$