Help me

I have 2 question

1.The graph of the quadratic $y = ax^2 + bx + c$ has the following properties: (1) The maximum value of $y = ax^2 + bx + c$ is 5, which occurs at $x = 3$. (2) The graph passes through the point $(0,-13)$. If the graph passes through the point $(4,m)$, then what is the value of $m$?

2.What is the area of the region defined by the equation $x^2+y^2 - 7 = 4y-14x+3$?

1. The vertex  (3, 5)  and the point  (0, - 13)  are on the graph ....and  (3,5)  is tthe vertex

The x coordinate of the vertex is given by :

-b / [2a]  =  3

-b = 6a

b = - 6a

And  since  (0, - 13)  is on the graph, then  c  =  -13

And using the vertex, we can find "a"  thusly

5  =  a(3)^2  - 6a(3)   - 13

18  = 9a- 18a

18  =  -9a

-2  = a       ⇒   b  = -6(-2)  =  12

So...our function is

y   =  -2x^2 + 12x - 13

And when x  =  4, m  =

-2(4)^2  + 12(4)  - 13

-32 + 48  - 13

3

So....the point (4,3)  is on the graph

Here's a graph : https://www.desmos.com/calculator/jhmml2xcki

2.What is the area of the region defined by the equation $x^2+y^2 - 7 = 4y-14x+3$?

Rewrite  as

x^2 + 14x +y^2 - 4y = 10

This is a circle....we need to complete the square on x and y  [ I'm assuming that you are familiar with this?? ]

x^2 + 14x + 49  + y^2 - 4y + 4  =  10 + 49 + 4

The right side simplifies to  63  = ( radius of the circle)^2

So... the area  is    pi * r^2   =  pi *63  ≈  197.9  units^2