Find the quadratic function whose graph satisfies the following.Write the function in the form f(x)=ax^2+bx+c.

Question

1. Vertex at (-4, 3), containing (-6, 11).
2. Vertex at (2, -5), containing (3, 1).
3. Vertex at (3, -1), containing (0, -6).
pls show me the complete solution of getting the a,b, and c....tnx!

Answer 1

Use the form

y = a(x-h)^2 + k

where (h,k) is the coordinate of the vertex to get started. Then multiply everthing out to get it in standard form.

PROBLEM NUMBER 1

y = a( x - (-4) ) ^2 + 3

simplifies to

y=a(x+4)^2 + 3

has to pass through the point (-6,11). Substitute in x= -6, y=11 and solve for a.

11=a(-6+4)^2 + 3
11=a(-2)^2+3
11=4a+3
8=4a
a=2

so we have
y = 2(x+4)^2 + 3

That's the equation in vertex form but you want it in standard from so we just mulitply it out and combine like terms.

y = 2( x^2+8x+16) + 3
y=2x^2+16x+32+3

finally giving us the answer to #1
y = 2 x^2 + 16x + 35

PROBLEM NUMBER 2

y = a(x-2)^2 + (-5)
y = a(x-2)^2 - 5

substitute in x=3 and y=1
1 = a(3-2)^2 - 5
1 = a - 5
a = 6

y = 6 (x-2)^2 -5

multiply out and simplify

y = 6(x^2-4x+4) - 5
y = 6 x^2 - 24x + 24 - 5

giving us the answer
y = 6 x^2 - 24x + 19

PROBLEM NUMBER 3

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

substituting in x=0 and y= -6
-6 = a (0-3)^2 - 1
-6 = a (-3)^2 - 1
-6 = 9a - 1
- 5 = 9a
a = - 5 / 9

y = - (5/9) (x-3)^2 - 1

multiplying out and simplifying

y = - (5/9) (x^2 - 6x + 9) - 1
y = - (5/9) x^2 + (10/3) x - 5 - 1

to give us a final answer of

y = - (5/9) x^2 + (10/3) x - 6

Good Luck!

Answer 2

1. If y = ax^2 + bx + c, then y' = 2ax + b. The x-coordinate of the vertex occurs when y' = 0, so 0 = 2a(-4) + b, or b = 8a. We now know the equation is of the form y = ax^2 + 8ax + c.

Since y = 3 when x = -4, we have 3 = 16a - 32a + c, so c = 3 + 16a, and the equation has the form y = ax^2 + 8ax + 3 + 16a. Finally, using the point (-6,11), we have 11 = 36a -48a + 3 + 16a, so a = 2. Therefore b = 16 and c = 35, and y = 2x^2 + 16x + 35.

The other parts can be solved in a similar manner.

Answer 3

replace each and every factor on y=ax^2+bx+c, because of the fact the factors fulfill the equation. 21=4a-2b+c (one million) 6=a+b+c (2) 26=9a+3b+c (3) So, you have simultaneous equations to remedy Take 2 pairs of them and do away with the comparable letter each and every time. From (one million) and (2), by skill of removing b, you have 33=6a+3c (call this as (4)) and from (one million) and (3) by skill of removing b lower back, you have one hundred fifteen=30a+5c (call this as (5)) remedy (4) and (5) and you discover a=3. replace a=3 on (4) and you have c=5 replace a=3 and c=5 on all of us of the preliminary 3 equations, we could say on (2) and you have b=-2 So, the quadratic function is f(x)=3x^2-2x+5

Answer 4

u cannot find 3 variables using 2 inputs.