QuestionThe graph of $y=4-x^{2}$ is: a) down parabola at $(0,4)$ , b) down at $(4,0)$ , c) up at $(0,4)$ , d) up at $(4,0)$ , e) circle radius 2.

Studdy Solution

STEP 1

Assumptions1. The equation given is $y=4-x^{}$ . . We need to determine the shape of the graph and the vertex of the graph.

STEP 2

The given equation is in the form of a quadratic equation. A quadratic equation is of the form $y=ax^{2}+bx+c$ . Here, $a=-1$ , $b=0$ , and $c=4$ .

STEP 3

The coefficient of $x^{2}$ is negative, which means the parabola opens downwards. This eliminates options c) and d) which suggest the parabola opens upwards.

STEP 4

The vertex of a parabola given by the equation $y=ax^{2}+bx+c$ is at the point $(-\frac{b}{2a},c-\frac{b^{2}}{4a})$ .

STEP 5

Substitute $a=-1$ and $b=0$ into the vertex formula to find the vertex of the parabola.
$Vertex = \left(-\frac{b}{2a},c-\frac{b^{2}}{4a}\right) = \left(-\frac{0}{2(-1)},4-\frac{0^{2}}{4(-1)}\right)$

STEP 6

Calculate the vertex of the parabola.
$Vertex = \left(0,4\right)$

STEP 7

The vertex of the parabola is at $(0,4)$ and it opens downwards. This matches option a) a downward opening parabola with vertex $(x, y)=(0,4)$ .
So, the correct answer is a) a downward opening parabola with vertex $(x, y)=(0,4)$ .

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QuestionThe graph of $y=4-x^{2}$ is: a) down parabola at $(0,4)$ , b) down at $(4,0)$ , c) up at $(0,4)$ , d) up at $(4,0)$ , e) circle radius 2.

Studdy Solution

STEP 1

Assumptions1. The equation given is $y=4-x^{}$ . . We need to determine the shape of the graph and the vertex of the graph.

STEP 2

The given equation is in the form of a quadratic equation. A quadratic equation is of the form $y=ax^{2}+bx+c$ . Here, $a=-1$ , $b=0$ , and $c=4$ .

STEP 3

The coefficient of $x^{2}$ is negative, which means the parabola opens downwards. This eliminates options c) and d) which suggest the parabola opens upwards.

STEP 4

The vertex of a parabola given by the equation $y=ax^{2}+bx+c$ is at the point $(-\frac{b}{2a},c-\frac{b^{2}}{4a})$ .

STEP 5

Substitute $a=-1$ and $b=0$ into the vertex formula to find the vertex of the parabola.
$Vertex = \left(-\frac{b}{2a},c-\frac{b^{2}}{4a}\right) = \left(-\frac{0}{2(-1)},4-\frac{0^{2}}{4(-1)}\right)$

STEP 6

Calculate the vertex of the parabola.
$Vertex = \left(0,4\right)$

STEP 7

The vertex of the parabola is at $(0,4)$ and it opens downwards. This matches option a) a downward opening parabola with vertex $(x, y)=(0,4)$ .
So, the correct answer is a) a downward opening parabola with vertex $(x, y)=(0,4)$ .

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QuestionThe graph of y=4−x2y=4-x^{2}y=4−x2 is: a) down parabola at (0,4)(0,4)(0,4), b) down at (4,0)(4,0)(4,0), c) up at (0,4)(0,4)(0,4), d) up at (4,0)(4,0)(4,0), e) circle radius 2.

Studdy Solution

STEP 1

Assumptions1. The equation given is y=4−xy=4-x^{}y=4−x. . We need to determine the shape of the graph and the vertex of the graph.

STEP 2

The given equation is in the form of a quadratic equation. A quadratic equation is of the form y=ax2+bx+cy=ax^{2}+bx+cy=ax2+bx+c. Here, a=−1a=-1a=−1, b=0b=0b=0, and c=4c=4c=4.

STEP 3

The coefficient of x2x^{2}x2 is negative, which means the parabola opens downwards. This eliminates options c) and d) which suggest the parabola opens upwards.

STEP 4

The vertex of a parabola given by the equation y=ax2+bx+cy=ax^{2}+bx+cy=ax2+bx+c is at the point (−b2a,c−b24a)(-\frac{b}{2a},c-\frac{b^{2}}{4a})(−2ab​,c−4ab2​).

STEP 5

STEP 6

Calculate the vertex of the parabola.Vertex=(0,4)Vertex = \left(0,4\right)Vertex=(0,4)

STEP 7

The vertex of the parabola is at (0,4)(0,4)(0,4) and it opens downwards. This matches option a) a downward opening parabola with vertex (x,y)=(0,4)(x, y)=(0,4)(x,y)=(0,4).So, the correct answer is a) a downward opening parabola with vertex (x,y)=(0,4)(x, y)=(0,4)(x,y)=(0,4).

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QuestionThe graph of $y=4-x^{2}$ is: a) down parabola at $(0,4)$ , b) down at $(4,0)$ , c) up at $(0,4)$ , d) up at $(4,0)$ , e) circle radius 2.

Assumptions1. The equation given is $y=4-x^{}$ . . We need to determine the shape of the graph and the vertex of the graph.

The given equation is in the form of a quadratic equation. A quadratic equation is of the form $y=ax^{2}+bx+c$ . Here, $a=-1$ , $b=0$ , and $c=4$ .

The coefficient of $x^{2}$ is negative, which means the parabola opens downwards. This eliminates options c) and d) which suggest the parabola opens upwards.

The vertex of a parabola given by the equation $y=ax^{2}+bx+c$ is at the point $(-\frac{b}{2a},c-\frac{b^{2}}{4a})$ .

Calculate the vertex of the parabola.
$Vertex = \left(0,4\right)$

The vertex of the parabola is at $(0,4)$ and it opens downwards. This matches option a) a downward opening parabola with vertex $(x, y)=(0,4)$ .
So, the correct answer is a) a downward opening parabola with vertex $(x, y)=(0,4)$ .