a parabola has its vertex at 1,1 and passes through the point -2,-2. Find the equation of this parabola

Mathematics

1 answer:

Jet001 [13]3 years ago

5 0

Answer:

The vertex form of the equation is $y = -\frac{1}{3}( x-1)^{2} +1$ .

Step-by-step explanation:

You will use the equation $y=a(x-h)^{2} +k$ to solve this parabola. (h, k) is the vertex. So, we plug this into the equation to get $y= a(x - 1)^{2} +1$ .

Then, we substitute the point as our x and y to get $-2 = 9a+1$ (a lot of simplifying was done here). Then, add 2 to the right side of the equation and isolate the $9a$ to get $9a = 3$ . Finally, divide by 9 on both sides to get $a=\frac{1}{3}$ .

Now, substitute your $a$ back into the equation to get y = $-\frac{1}{3}(x-1)^{2}+1$ .

This is in vertex form. If the answer is needed in standard form, simply distribute and simplify to get $y = -\frac{1}{3} x^{2} +\frac{2}{3} x+\frac{2}{3}$ .

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Montano1993 [528]

<h3>Answer: 7.1</h3>

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Explanation:

The two points are (x1,y1) = (1,3) and (x2,y2) = (8,4)

Use the distance formula to get,

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (3-4)^2}\\\\d = \sqrt{(-7)^2 + (-1)^2}\\\\d = \sqrt{49 + 1}\\\\d = \sqrt{50}\\\\d = \sqrt{25*2}\\\\d = \sqrt{25}*\sqrt{2}\\\\d = 5\sqrt{2}\\\\d \approx 7.0710678\\\\d \approx 7.1$

Alternatively, you can plot the two points A(1,3) and B(8,4) on the same xy grid. Then plot point C(8,3). Note that C has the same x coordinate as B, and the same y coordinate as A. Triangle ABC is a right triangle, where the 90 degree angle is at point C. The distance from A to B is the same as finding the length of AB, which is the hypotenuse.

This means we can apply the pythagorean theorem to find the hypotenuse AB. The distance formula is a modified version of the pythagorean theorem. Refer to the diagram below.