Math / Algebra

QuestionIdentify the vertex, and the $y$ -intercept then graph $f(x)=x^{2}-4 x+2$ . Hint: Click the vertex, then click another point of the parabola (like the $y$ -intercept). The vertex is ( $\square$ $\square$ The $y$ -intercept is ( 0 , $\square$ )

Studdy Solution

STEP 1

1. The function $f(x) = x^2 - 4x + 2$ is a quadratic function.
2. The vertex form of a quadratic function is $f(x) = a(x-h)^2 + k$ , where $(h, k)$ is the vertex.
3. The $y$ -intercept occurs when $x = 0$ .

STEP 2

1. Identify the vertex of the quadratic function.
2. Identify the $y$ -intercept of the quadratic function.
3. Graph the quadratic function using the vertex and $y$ -intercept.

STEP 3

To find the vertex of the quadratic function $f(x) = x^2 - 4x + 2$ , we can use the vertex formula for a parabola in standard form $ax^2 + bx + c$ . The $x$ -coordinate of the vertex is given by:
$x = -\frac{b}{2a}$
For $f(x) = x^2 - 4x + 2$ , $a = 1$ and $b = -4$ .

STEP 4

Calculate the $x$ -coordinate of the vertex:
$x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$

STEP 5

Substitute $x = 2$ back into the function to find the $y$ -coordinate of the vertex:
$f(2) = (2)^2 - 4(2) + 2 = 4 - 8 + 2 = -2$
Thus, the vertex is $(2, -2)$ .

STEP 6

To find the $y$ -intercept, substitute $x = 0$ into the function:
$f(0) = (0)^2 - 4(0) + 2 = 2$
Thus, the $y$ -intercept is $(0, 2)$ .

STEP 7

To graph the quadratic function $f(x) = x^2 - 4x + 2$ , plot the vertex $(2, -2)$ and the $y$ -intercept $(0, 2)$ . Then, draw a symmetric parabola opening upwards, as the coefficient of $x^2$ is positive.
The vertex is $(2, -2)$ and the $y$ -intercept is $(0, 2)$ .

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QuestionIdentify the vertex, and the $y$ -intercept then graph $f(x)=x^{2}-4 x+2$ . Hint: Click the vertex, then click another point of the parabola (like the $y$ -intercept). The vertex is ( $\square$ $\square$ The $y$ -intercept is ( 0 , $\square$ )

Studdy Solution

STEP 1

1. The function $f(x) = x^2 - 4x + 2$ is a quadratic function.
2. The vertex form of a quadratic function is $f(x) = a(x-h)^2 + k$ , where $(h, k)$ is the vertex.
3. The $y$ -intercept occurs when $x = 0$ .

STEP 2

1. Identify the vertex of the quadratic function.
2. Identify the $y$ -intercept of the quadratic function.
3. Graph the quadratic function using the vertex and $y$ -intercept.

STEP 3

To find the vertex of the quadratic function $f(x) = x^2 - 4x + 2$ , we can use the vertex formula for a parabola in standard form $ax^2 + bx + c$ . The $x$ -coordinate of the vertex is given by:
$x = -\frac{b}{2a}$
For $f(x) = x^2 - 4x + 2$ , $a = 1$ and $b = -4$ .

STEP 4

Calculate the $x$ -coordinate of the vertex:
$x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$

STEP 5

Substitute $x = 2$ back into the function to find the $y$ -coordinate of the vertex:
$f(2) = (2)^2 - 4(2) + 2 = 4 - 8 + 2 = -2$
Thus, the vertex is $(2, -2)$ .

STEP 6

To find the $y$ -intercept, substitute $x = 0$ into the function:
$f(0) = (0)^2 - 4(0) + 2 = 2$
Thus, the $y$ -intercept is $(0, 2)$ .

STEP 7

To graph the quadratic function $f(x) = x^2 - 4x + 2$ , plot the vertex $(2, -2)$ and the $y$ -intercept $(0, 2)$ . Then, draw a symmetric parabola opening upwards, as the coefficient of $x^2$ is positive.
The vertex is $(2, -2)$ and the $y$ -intercept is $(0, 2)$ .

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QuestionIdentify the vertex, and the yyy-intercept then graph f(x)=x2−4x+2f(x)=x^{2}-4 x+2f(x)=x2−4x+2. Hint: Click the vertex, then click another point of the parabola (like the yyy-intercept). The vertex is ( □\square□ □\square□ The yyy-intercept is ( 0 , □\square□ )

Studdy Solution

STEP 1

1. The function f(x)=x2−4x+2 f(x) = x^2 - 4x + 2 f(x)=x2−4x+2 is a quadratic function.2. The vertex form of a quadratic function is f(x)=a(x−h)2+k f(x) = a(x-h)^2 + k f(x)=a(x−h)2+k, where (h,k)(h, k)(h,k) is the vertex.3. The y y y-intercept occurs when x=0 x = 0 x=0.

STEP 2

1. Identify the vertex of the quadratic function.2. Identify the y y y-intercept of the quadratic function.3. Graph the quadratic function using the vertex and y y y-intercept.

STEP 3

STEP 4

Calculate the x x x-coordinate of the vertex:x=−−42×1=42=2 x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2 x=−2×1−4​=24​=2

STEP 5

Substitute x=2 x = 2 x=2 back into the function to find the y y y-coordinate of the vertex:f(2)=(2)2−4(2)+2=4−8+2=−2 f(2) = (2)^2 - 4(2) + 2 = 4 - 8 + 2 = -2 f(2)=(2)2−4(2)+2=4−8+2=−2Thus, the vertex is (2,−2)(2, -2)(2,−2).

STEP 6

To find the y y y-intercept, substitute x=0 x = 0 x=0 into the function:f(0)=(0)2−4(0)+2=2 f(0) = (0)^2 - 4(0) + 2 = 2 f(0)=(0)2−4(0)+2=2Thus, the y y y-intercept is (0,2)(0, 2)(0,2).

STEP 7

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QuestionIdentify the vertex, and the $y$ -intercept then graph $f(x)=x^{2}-4 x+2$ . Hint: Click the vertex, then click another point of the parabola (like the $y$ -intercept). The vertex is ( $\square$ $\square$ The $y$ -intercept is ( 0 , $\square$ )

1. The function $f(x) = x^2 - 4x + 2$ is a quadratic function.
2. The vertex form of a quadratic function is $f(x) = a(x-h)^2 + k$ , where $(h, k)$ is the vertex.
3. The $y$ -intercept occurs when $x = 0$ .

1. Identify the vertex of the quadratic function.
2. Identify the $y$ -intercept of the quadratic function.
3. Graph the quadratic function using the vertex and $y$ -intercept.

Calculate the $x$ -coordinate of the vertex:
$x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$

Substitute $x = 2$ back into the function to find the $y$ -coordinate of the vertex:
$f(2) = (2)^2 - 4(2) + 2 = 4 - 8 + 2 = -2$
Thus, the vertex is $(2, -2)$ .

To find the $y$ -intercept, substitute $x = 0$ into the function:
$f(0) = (0)^2 - 4(0) + 2 = 2$
Thus, the $y$ -intercept is $(0, 2)$ .