Quadratic minimum x-value Practice Questions (Grade 9)

Question 1easy

The equation below defines $y$ as a function of $x$ . At what value of $x$ is $y$ smallest?

$y = x^2 - 6x + 5$

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 1$ and $b = -6$ , so $x = -\dfrac{-6}{2(1)} = \dfrac{6}{2} = 3$ . Therefore, $y$ is smallest when $x = 3$ .

Question 2easy

The equation $y = x^2 + 8x + 7$ relates $x$ and $y$ . What value of $x$ gives the least value of $y$ ?

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 1$ and $b = 8$ , so $x = -\dfrac{8}{2(1)} = -\dfrac{8}{2} = -4$ . Therefore, $y$ is smallest when $x = -4$ .

Question 3easy

A parabola opens upward and is described by $y = x^2 - 18x + 65$ . At what $x$ -coordinate does $y$ attain its minimum?

Because this parabola opens upward, the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 1$ and $b = -18$ , so $x = -\dfrac{-18}{2(1)} = \dfrac{18}{2} = 9$ . Therefore, $y$ reaches its minimum when $x = 9$ .

Question 4easy

The graph of $y = x^2 - 4x + 1$ is a parabola that opens upward. What is the $x$ -coordinate of its lowest point?

Because this parabola opens upward, its lowest point is the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 1$ and $b = -4$ , so $x = -\dfrac{-4}{2(1)} = \dfrac{4}{2} = 2$ . Therefore, the lowest point has $x$ -coordinate $2$ .

Question 5easy

The equation $y = x^2 + 4x + 3$ models a quantity $y$ in terms of $x$ . For what value of $x$ is $y$ as small as possible?

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 1$ and $b = 4$ , so $x = -\dfrac{4}{2(1)} = -\dfrac{4}{2} = -2$ . Therefore, $y$ is as small as possible when $x = -2$ .

Question 6medium

For what value of $x$ does $y = 2x^2 + 12x + 4$ reach a minimum?

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 2$ and $b = 12$ , so $x = -\dfrac{12}{2(2)} = -\dfrac{12}{4} = -3$ . Therefore, $y$ reaches its minimum when $x = -3$ .

Question 7medium

Given $y = 4x^2 - 8x + 3$ , at what value of $x$ is $y$ minimized?

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 4$ and $b = -8$ , so $x = -\dfrac{-8}{2(4)} = \dfrac{8}{8} = 1$ . Therefore, $y$ is minimized when $x = 1$ .

Question 8medium

For the relation $y = 3x^2 - 30x + 70$ , find the value of $x$ at which $y$ is smallest.

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$ , the $x$ -value at the vertex is $x = -\dfrac{b}{2a}$ . Here $a = 3$ and $b = -30$ , so $x = -\dfrac{-30}{2(3)} = \dfrac{30}{6} = 5$ . Therefore, $y$ is smallest when $x = 5$ .

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