Question

1st question answer pls

Answer 1

y = -3(x + 1)^2 + 2

Explanation:

y = a(x - h)^2 + k is the vertex form of a quadratic, where

• (x, y) are any point that lies on the parabola,
• a is a constant determining whether the parabola opens upward or downward,
• and (h, k) are the coordinates of the vertex.

Finding (h, k):

We see from the graph that the vertex is a maximum and its coordinates are (-1, 2). Thus h is -1 and k is 2. Since h becomes negative, it will be 1 in the parentheses: (x - (-1) = (x + 1).

Finding a:

In order to find a, we will need to plug in a point on the parabola for (x, y) and (-1, 2) for h and k. We see that (0, -1) lies on the parabola so we can use this point for (x, y).

-1 = a(0 - (-1))^2 + 2

-1 = a(0 + 1)^2 + 2

-3 = a(1)^2

-3 = a

Thus, a = -3.

Thus, the exact equation in vertex form of the parabola is:

y = -3(x + 1)^2 + 2

I attached a picture from Desmos Graphing Calculator that shows how the equation I provided works and contains the two points you marked on the parabola, including (-1, 2) aka the maximum, and (0, -1) aka the y-intercept.

Answer 2

let's take a peek at the picture above, hmmm let's notice the vertex is at (-1 , 2), now let's get a point besides the vertex hmmm let's see it passes through (-2 , -1).

So we can reword that as what's the equation of a quadratic whose vertex is at (-1 , 2) and it passes through (-2 , -1)?

`~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{a~is~negative}{op ens~\cap}\qquad \stackrel{a~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill`

`\begin{cases} h=-1\\ k=2\\ \end{cases}\implies y=a(~~x-(-1)~~)^2 + 2\hspace{4em}\textit{we also know that} \begin{cases} x=-2\\ y=-1 \end{cases} \\\\\\ -1=a( ~~-2-(-1) ~~ )^2 + 2\implies -3=a(-2+1)^2\implies -3=a \\\\\\ ~\hfill~ {\Large \begin{array}{llll} y=-3(x+1)^2 + 2 \end{array}} ~\hfill~`