Question

consider the parabola of equation y=(a-3)x^2-2(a+1)x+a-1 with a E R. Determine for which values of a this parabola: 1) does not intersect the x axis at any point; 2) has the vertex with a negative abscissa; 3) has the concavity facing downwards; 4) passes through point P (-2; 4).

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Answer to a math question consider the parabola of equation y=(a-3)x^2-2(a+1)x+a-1 with a E R. Determine for which values of a this parabola: 1) does not intersect the x axis at any point; 2) has the vertex with a negative abscissa; 3) has the concavity facing downwards; 4) passes through point P (-2; 4).

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Hester
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115 Answers
1) Per determinare per quali valori di a la parabola non interseca l'asse x in nessun punto, dobbiamo considerare il discriminante della funzione quadratica. Se il discriminante è negativo, la parabola non interseca l'asse x.

Il discriminante è dato da:

\Delta = b^2 - 4ac

Dove, nell'equazione y = (a-3)x^2 - 2(a+1)x + a - 1, abbiamo a = (a-3), b = -2(a+1), e c = a-1.

Sostituendo questi valori nell'equazione del discriminante otteniamo:

\Delta = (-2(a+1))^2 - 4(a-3)(a-1)

Espandendo e semplificando otteniamo:

\Delta = 4(a^2 + 2a + 1) - 4(a^2 - 4a + 3)

\Delta = 4a^2 + 8a + 4 - 4a^2 + 16a - 12

\Delta = 24a - 8

Perché la parabola non intersechi l'asse x in nessun punto, il discriminante deve essere negativo, quindi:

24a - 8 < 0

24a < 8

a < \frac{1}{3}

Quindi, la parabola non interseca l'asse x in nessun punto per a < \frac{1}{3}.

2) Per determinare per quali valori di a la parabola ha il vertice con ascissa negativa, dobbiamo trovare l'ascissa del vertice della parabola. L'ascissa del vertice di una parabola di equazione y = ax^2 + bx + c è data da x = -\frac{b}{2a}.

Nel nostro caso, l'ascissa del vertice è:

x = -\frac{-2(a+1)}{2(a-3)} = \frac{a+1}{a-3}

Per fare in modo che l'ascissa del vertice sia negativa, dobbiamo risolvere l'inequazione:

\frac{a+1}{a-3} < 0

La quale dà come soluzione:

-1 < a < 3

Quindi, la parabola ha il vertice con ascissa negativa per -1 < a < 3.

3) Per determinare per quali valori di a la parabola ha la concavità rivolta verso il basso, dobbiamo considerare il coefficiente del termine x^2, che è a-3. Per avere la concavità rivolta verso il basso, il coefficiente a-3 deve essere negativo, quindi:

a - 3 < 0

a < 3

Quindi, la parabola ha la concavità rivolta verso il basso per a < 3.

4) Per determinare per quali valori di a la parabola passa per il punto P(-2, 4), dobbiamo sostituire le coordinate x = -2 e y = 4 nell'equazione della parabola e risolvere per a. Quindi abbiamo:

4 = (a-3)(-2)^2 - 2(a+1)(-2) + a - 1

4 = 4(a-3) + 4(a+1) + a - 1

4 = 4a - 12 + 4a + 4 + a - 1

4 = 9a - 9

9 = 9a

a = 1

Quindi, la parabola passa per il punto P(-2, 4) quando a = 1.

**Risposta:**
1) La parabola non interseca l'asse x per a < \frac{1}{3}.
2) La parabola ha il vertice con ascissa negativa per -1 < a < 3.
3) La parabola ha la concavità rivolta verso il basso per a < 3.
4) La parabola passa per il punto P(-2, 4) quando a = 1.

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