Question

Determine the quadratic equation for the parabola:
minimum value: -5; x-intercepts: -2 and 3

Answer 1

Please help! So stuck!

Answer 2

y=a(x-3)(x-(-2))

so
put axis of symmetry into it
and find a
y=-5 when put x of axis of symmetry

Answer 3

I don't understand :(

Answer 4

y=a(x-x1)(x-x2)

axis of symmetry is
x1 +x2
x= --------
2

Answer 5

I have to get the answers like this,
f(x) = 4/5x2 + 2/5x - 48/5 f(x) = 2x2 + 4x - 48
f(x) = 2/5x2 + 4/5x - 48/5 f(x) = 5x2 + 4x - 48

Answer 6

These are the answer im given and I am not getting how to get these

Answer 7

x1=-2
x2=+3
so
axis of symmetry is
-2+3 1
x=------=----
2 2
so put x=1/2
to y=a(x+2)(x-3)
this is minimum value
-5=a(1/2+2)(1/2-3)
-5=a(5/2)(-5/2)
so a=4/5

Answer 8

can you go on ?

Answer 9

Do I do 4/5(5/2(-5/2) and distribute?

Answer 10

yes solve for a

Answer 11

Doesn't make sense I got 2 and -2

Answer 12

\[y=a(x+2)(x-3) \rightarrow put\x=\frac{ 1 }{2 } \\-5=a(\frac{ 1 }{2 }+2)(\frac{ 1 }{2 }-3) \\-5=a(\frac{ 5 }{2 })(\frac{ 1-6 }{2 })\\-5=a(\frac{ 5 }{2 })(\frac{ -5 }{2 })\\-5=a(\frac{- 25 }{4 })\\ \frac{ -5 }{ 1 }=a(\frac{- 25 }{4 })\\ \frac{ -5*4 }{ -25 }=a\\a=\frac{ 4 }{ 5 }\]