Question

Sa se determine functia de gradul 2 f:R->R, f(x)=ax^2 +bx+c, a≠0, daca graficul este tangent axei 0x si intersecteaza dreapta y-2x+12=0 in punctele de abscise x=-2, x=4

Answer 1

Răspuns:

$f(x) = -1.5x^2 + 5x$

Explicație pas cu pas:

y = 2x - 12

x = -2

y = -4 -12 = -16

x = 4

y = 8 - 12 = -4

a < 0

$\Delta = 0$

$b^2 - 4ac = 0$

4a - 2b + c = -16

8a - 4b + 2c = -32

16a + 4b + c = -4

24a + 3c = -36

3(8a + c) = -36

8a + c = -12

c = -4(2a + 3)

$a\times \frac{b^2}{a^2} + b\times \frac{-b}{a} + c = 0$

$\frac{b^2 - b^2}{a} + c = 0$

$c = 0$

0 = -4(2a + 3)

2a + 3 = 0

2a = -3

a = -3/2 = -1.5

-6 -2b + 0 = -16

-2b = -10

b = 5

$f(x) = -1.5x^2 + 5x$