Hi, can anyone help with b) and c)?

steps below

Step-by-step explanation:

From the graph: parabola pass (0,4) (3,4) (1,2)

equation: y = ax² + bx + c

(0,4): c = 4

(3,4): 9a + 3b + 4 = 4       9a+3b = 0       3a + b = 0   ... (1)

(1,2):  a + b + 4 = 2            a+b = -2   ... (2)

(1)-(2): 2a = 2        a = 1

(2):      b = -3

parabola equation: f(x) = x² - 3x +4

(a)  f(3) = 3² - 3*3 + 4 = 4

(b) normal line: parallel to tangent line y = 3x -5 at (3,4)

    slope of line N: -1/3   pass (3,4)

    equation of N: (y-4)/(x-3) = -1/3         y-4 = -1/3 x + 1

     line N: 1/3 x + y - 5 = 0

(c) picture attached

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1