Peyton and Tanner were lifting a large boulder and relocating it within their property. Peyton ispushing the boulder with a forc
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Answer:
y = -3(x - 4)² - 2
Step-by-step explanation:
Given the vertex, (4, -2), and the point (2, -14):
We can use the vertex form of the quadratic equation:
y = a(x - h)² + k
Where:
(h, k) = vertex
a = determines whether the graph opens up or down, and it also makes the parent function <u>wider</u> or <u>narrower</u>.
- <u>positive</u> value of a = opens <u><em>upward</em></u>
- <u>negative</u> value of a = opens <u><em>downward</em></u>
- a is between 0 and 1, (0 < a < 1) the graph is <u><em>wider</em></u> than the parent function.
- a > 1, the graph is <u><em>narrower</em></u> than the parent function.
<em>h </em>=<em> </em>determines how far left or right the parent function is translated.
- h = positive, the function is translated <em>h</em> units to the right.
- h = negative, the function is translated |<em>h</em>| units to the left.
<em>k</em> determines how far up or down the parent function is translated.
- k = positive: translate <em>k</em> units <u><em>up</em></u>.
- k = negative, translate <em>k</em> units <u><em>down</em></u>.
Now that I've set up the definitions for each variable of the vertex form, we can determine the quadratic equation using the given vertex and the point:
vertex (h, k): (4, -2)
point (x, y): (2, -14)
Substitute these values into the vertex form to solve for a:
y = a(x - h)² + k
-14 = a(2 - 4)² -2
-14 = a (-2)² -2
-14 = a4 + -2
Add to to both sides:
-14 + 2 = a4 + -2 + 2
-12 = 4a
Divide both sides by 4 to solve for a:
-12/4 = 4a/4
-3 = a
Therefore, the quadratic equation inI vertex form is:
y = -3(x - 4)² - 2
The parabola is downward-facing, and is vertically compressed by a factor of -3. The graph is also horizontally translated 4 units to the right, and vertically translated 2 units down.
Attached is a screenshot of the graph where it shows the vertex and the given point, using the vertex form that I came up with.
Please mark my answers as the Brainliest, if you find this helpful :)
