Quadratic minimum from vertex form Practice Questions

Warm-upeasy

$f(x) = (x - 8)^2 + 1$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Here $h = 8$ , so the vertex is at $x = 8$ . Therefore, $f(x)$ reaches its minimum when $x = 8$ .

Mid-gameeasy

$f(x) = (x + 3)^2 - 6$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Rewrite $(x + 3)^2$ as $(x - (-3))^2$ : the vertex is at $x = -3$ . Therefore, $f(x)$ reaches its minimum when $x = -3$ .

Mid-gameeasy

$f(x) = (x - 2)^2 + 10$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Here $h = 2$ , so the vertex is at $x = 2$ . Therefore, $f(x)$ reaches its minimum when $x = 2$ .

Mid-gameeasy

$f(x) = (x - 9)^2 - 2$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Here $h = 9$ , so the vertex is at $x = 9$ . Therefore, $f(x)$ reaches its minimum when $x = 9$ .

Mid-gameeasy

$f(x) = 3(x - 1)^2 + 7$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 3 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Here $h = 1$ , so the vertex is at $x = 1$ . The leading coefficient $3$ does not change the $x$ -coordinate of the vertex. Therefore, $f(x)$ reaches its minimum when $x = 1$ .

Mid-gameeasy

$f(x) = (x + 6)^2 + 8$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Rewrite $(x + 6)^2$ as $(x - (-6))^2$ : the vertex is at $x = -6$ . Therefore, $f(x)$ reaches its minimum when $x = -6$ .

Mid-gameeasy

$f(x) = (x - 5)^2 + 11$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Here $h = 5$ , so the vertex is at $x = 5$ . Therefore, $f(x)$ reaches its minimum when $x = 5$ .

Buzzer beatereasy

$f(x) = 4(x + 1)^2 - 3$

The function $f$ is defined by the given equation. For what value of $x$ does $f(x)$ reach its minimum?

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 4 > 0$ , so the graph opens upward and the minimum occurs at the vertex. Rewrite $(x + 1)^2$ as $(x - (-1))^2$ : the vertex is at $x = -1$ . The leading coefficient $4$ does not change the $x$ -coordinate of the vertex. Therefore, $f(x)$ reaches its minimum when $x = -1$ .

Quadratic minimum from vertex form. Game on.

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