Find the quadratic function y=f(x) whose graph has a vertex (−3,4) and passes through the point (−7,0). write the function in standard form.

Step-by-step explanation:

This is a parabola since a quadratic is a parabola.  The standard form for a parabola is y = ax² + bx + c

but before we do that, we will use the vertex form, since it will make our work easier at the beginning.  

First and foremost, when we plot the vertex and the given point, the vertex is higher up than is the point; that means that this parabola opens upside down, and its vertex form will be

y=-|a|(x - h)² + k

The absolute value is out in front of the a, so we know that the value of a is positive, but the quadratic itself is negative (upside down) and we will find that math takes care of that negative that needs to be out front.  So we need to solve for a by filling in the x, y, h, and k values from the point and the vertex:  x = -7, y = 0, h = -3, k = 4

0 = a(-7 - (-3))² + 4 and

0 = a(-7 + 3)² + 4 and

0 = a(-4)² + 4 and

0 = a(16) + 4 and

0 = 16a + 4 and

-4 = 16a so

Now that we know a, we can plug it back into the vertex form and then put it into standard form from there.

Now we will FOIL out what's inside the parenthesis to get

Simplify by distributing the -1/4 into the parenthesis:

Combine like terms to get

And there you go!

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.