Math / Algebra

QuestionConsider the parabola given by the equation: $y=-1 x^{2}+2 x+3$
Find the following for this parabola: A) The vertex $=($ $\square$ $\square$ ) B) The $y$ intercept is the point ( 0 , $\square$ ) C) Find the two values of $x$ that correspond to the $x$ intercepts of the parabola and write them as a list, separated by commas: $x=$ $\square$

Studdy Solution

STEP 1

1. The given parabola is in the standard form $y = ax^2 + bx + c$ .
2. The vertex of a parabola in standard form can be found using the formula $x = -\frac{b}{2a}$ .
3. The y-intercept occurs where $x = 0$ .
4. The x-intercepts occur where $y = 0$ .

STEP 2

1. Identify the coefficients $a$ , $b$ , and $c$ .
2. Calculate the vertex of the parabola.
3. Determine the y-intercept.
4. Solve for the x-intercepts.

STEP 3

Identify the coefficients from the equation $y = -1x^2 + 2x + 3$ .
Here, $a = -1$ , $b = 2$ , and $c = 3$ .

STEP 4

Calculate the x-coordinate of the vertex using the formula $x = -\frac{b}{2a}$ .
$x = -\frac{2}{2(-1)} = 1$

STEP 5

Substitute $x = 1$ back into the equation to find the y-coordinate of the vertex.
$y = -1(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4$
Thus, the vertex is $(1, 4)$ .

STEP 6

Determine the y-intercept by setting $x = 0$ in the equation.
$y = -1(0)^2 + 2(0) + 3 = 3$
The y-intercept is the point $(0, 3)$ .

STEP 7

Solve for the x-intercepts by setting $y = 0$ and solving the quadratic equation.
$0 = -1x^2 + 2x + 3$
Rearrange to:
$-x^2 + 2x + 3 = 0$

STEP 8

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the x-intercepts.
$x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(3)}}{2(-1)}$ $x = \frac{-2 \pm \sqrt{4 + 12}}{-2}$ $x = \frac{-2 \pm \sqrt{16}}{-2}$ $x = \frac{-2 \pm 4}{-2}$
Calculate the two solutions:
$x_1 = \frac{-2 + 4}{-2} = -1$ $x_2 = \frac{-2 - 4}{-2} = 3$
The x-intercepts are $x = -1, 3$ .
A) The vertex is $(1, 4)$ .
B) The y-intercept is the point $(0, 3)$ .
C) The x-intercepts are $x = -1, 3$ .

Was this helpful?

Studdy Solution

STEP 1

1. The given parabola is in the standard form $y = ax^2 + bx + c$ .
2. The vertex of a parabola in standard form can be found using the formula $x = -\frac{b}{2a}$ .
3. The y-intercept occurs where $x = 0$ .
4. The x-intercepts occur where $y = 0$ .

STEP 2

1. Identify the coefficients $a$ , $b$ , and $c$ .
2. Calculate the vertex of the parabola.
3. Determine the y-intercept.
4. Solve for the x-intercepts.

STEP 3

Identify the coefficients from the equation $y = -1x^2 + 2x + 3$ .
Here, $a = -1$ , $b = 2$ , and $c = 3$ .

STEP 4

Calculate the x-coordinate of the vertex using the formula $x = -\frac{b}{2a}$ .
$x = -\frac{2}{2(-1)} = 1$

STEP 5

Substitute $x = 1$ back into the equation to find the y-coordinate of the vertex.
$y = -1(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4$
Thus, the vertex is $(1, 4)$ .

STEP 6

Determine the y-intercept by setting $x = 0$ in the equation.
$y = -1(0)^2 + 2(0) + 3 = 3$
The y-intercept is the point $(0, 3)$ .

STEP 7

Solve for the x-intercepts by setting $y = 0$ and solving the quadratic equation.
$0 = -1x^2 + 2x + 3$
Rearrange to:
$-x^2 + 2x + 3 = 0$

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. The given parabola is in the standard form y=ax2+bx+c y = ax^2 + bx + c y=ax2+bx+c.2. The vertex of a parabola in standard form can be found using the formula x=−b2a x = -\frac{b}{2a} x=−2ab​.3. The y-intercept occurs where x=0 x = 0 x=0.4. The x-intercepts occur where y=0 y = 0 y=0.

STEP 2

1. Identify the coefficients a a a, b b b, and c c c.2. Calculate the vertex of the parabola.3. Determine the y-intercept.4. Solve for the x-intercepts.

STEP 3

Identify the coefficients from the equation y=−1x2+2x+3 y = -1x^2 + 2x + 3 y=−1x2+2x+3.Here, a=−1 a = -1 a=−1, b=2 b = 2 b=2, and c=3 c = 3 c=3.

STEP 4

Calculate the x-coordinate of the vertex using the formula x=−b2a x = -\frac{b}{2a} x=−2ab​.x=−22(−1)=1 x = -\frac{2}{2(-1)} = 1 x=−2(−1)2​=1

STEP 5

Substitute x=1 x = 1 x=1 back into the equation to find the y-coordinate of the vertex.y=−1(1)2+2(1)+3=−1+2+3=4 y = -1(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4 y=−1(1)2+2(1)+3=−1+2+3=4Thus, the vertex is (1,4) (1, 4) (1,4).

STEP 6

Determine the y-intercept by setting x=0 x = 0 x=0 in the equation.y=−1(0)2+2(0)+3=3 y = -1(0)^2 + 2(0) + 3 = 3 y=−1(0)2+2(0)+3=3The y-intercept is the point (0,3) (0, 3) (0,3).

STEP 7

Solve for the x-intercepts by setting y=0 y = 0 y=0 and solving the quadratic equation.0=−1x2+2x+3 0 = -1x^2 + 2x + 3 0=−1x2+2x+3Rearrange to:−x2+2x+3=0 -x^2 + 2x + 3 = 0 −x2+2x+3=0

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The given parabola is in the standard form $y = ax^2 + bx + c$ .
2. The vertex of a parabola in standard form can be found using the formula $x = -\frac{b}{2a}$ .
3. The y-intercept occurs where $x = 0$ .
4. The x-intercepts occur where $y = 0$ .

1. Identify the coefficients $a$ , $b$ , and $c$ .
2. Calculate the vertex of the parabola.
3. Determine the y-intercept.
4. Solve for the x-intercepts.

Identify the coefficients from the equation $y = -1x^2 + 2x + 3$ .
Here, $a = -1$ , $b = 2$ , and $c = 3$ .

Calculate the x-coordinate of the vertex using the formula $x = -\frac{b}{2a}$ .
$x = -\frac{2}{2(-1)} = 1$

Substitute $x = 1$ back into the equation to find the y-coordinate of the vertex.
$y = -1(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4$
Thus, the vertex is $(1, 4)$ .

Determine the y-intercept by setting $x = 0$ in the equation.
$y = -1(0)^2 + 2(0) + 3 = 3$
The y-intercept is the point $(0, 3)$ .

Solve for the x-intercepts by setting $y = 0$ and solving the quadratic equation.
$0 = -1x^2 + 2x + 3$
Rearrange to:
$-x^2 + 2x + 3 = 0$