Question

consider the parabola given by the equation: (f(x) = 2x^2 - 8x - 13)

find the following for this parabola:

a) the vertex:

b) the vertical intercept is the point

c) find the coordinates of the two (x)-intercepts of the parabola and write them as a list, separated by commas:

it is ok to round your value(s) to to two decimal places.

question 4

put the equation (y = x^2 - 2x) into the form (y = (x - h)^2 + k):

answer: (y =)

Explanation:

🆕 New Concept Discovered: Properties of Quadratic Functions
Find the vertex, intercepts, and standard forms of parabolas.

Step 1: Find the vertex of \( f(x) = 2x^2 - 8x - 13 \)

For a quadratic function in standard form \( f(x) = ax^2 + bx + c \), the \( x \)-coordinate of the vertex is given by:

$$ x = -\frac{b}{2a} $$

Here, \( a = 2 \), \( b = -8 \), and \( c = -13 \):

$$ x = -\frac{-8}{2(2)} = \frac{8}{4} = 2 $$

Now, find the \( y \)-coordinate by evaluating \( f(2) \):

$$ f(2) = 2(2)^2 - 8(2) - 13 $$

$$ f(2) = 2(4) - 16 - 13 $$

$$ f(2) = 8 - 16 - 13 = -21 $$

The vertex is \( (2, -21) \).

Step 2: Find the vertical intercept

The vertical intercept (or \( y \)-intercept) occurs where \( x = 0 \):

$$ f(0) = 2(0)^2 - 8(0) - 13 = -13 $$

Written as a coordinate point, the vertical intercept is \( (0, -13) \).

Step 3: Find the \( x \)-intercepts

The \( x \)-intercepts occur where \( f(x) = 0 \):

$$ 2x^2 - 8x - 13 = 0 $$

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

$$ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(-13)}}{2(2)} $$

$$ x = \frac{8 \pm \sqrt{64 + 104}}{4} $$

$$ x = \frac{8 \pm \sqrt{168}}{4} $$

Calculating the decimal values:

$$ \sqrt{168} \approx 12.9615 $$

$$ x_1 = \frac{8 + 12.9615}{4} \approx \frac{20.9615}{4} \approx 5.24 $$

$$ x_2 = \frac{8 - 12.9615}{4} \approx \frac{-4.9615}{4} \approx -1.24 $$

Written as coordinate points: \( (5.24, 0), (-1.24, 0) \).

Step 4: Convert \( y = x^2 - 2x \) to the form \( y = (x - h)^2 + k \)

To rewrite the equation, we complete the square for the expression \( x^2 - 2x \):

1. Take half of the coefficient of \( x \): \( \frac{-2}{2} = -1 \).
2. Square this value: \( (-1)^2 = 1 \).
3. Add and subtract this value inside the equation:

$$ y = (x^2 - 2x + 1) - 1 $$

$$ y = (x - 1)^2 - 1 $$

Answer:

First Problem:

• A) The vertex: (2, -21)
• B) The vertical intercept is the point: (0, -13)
• C) The coordinates of the two x-intercepts: (5.24, 0), (-1.24, 0)

Question 4:

• Answer: y = (x - 1)^2 - 1