Question

determines for which values of k the parabola of equation y= x^2-2(k-3)x-k+15 has at least one point in common with the x axis and intersects the y axis at a point of positive ordinate

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Answer to a math question determines for which values of k the parabola of equation y= x^2-2(k-3)x-k+15 has at least one point in common with the x axis and intersects the y axis at a point of positive ordinate

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Esmeralda
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98 Answers
Per determinare i valori di k per cui la parabola interseca l'asse delle x e l'asse y nei punti descritti, dobbiamo considerare le condizioni:

1. La parabola interseca l'asse delle x se il discriminante della funzione quadratica è maggiore di zero.
2. La parabola interseca l'asse y in un punto di ordinata positiva se k è tale che il termine noto della parabola è positivo.

La parabola è definita dall'equazione y = x^2 - 2(k-3)x - k + 15 .

1. Per determinare quando la parabola interseca l'asse delle x, calcoliamo il discriminante:

Il discriminante della funzione quadratica è dato da \Delta = b^2 - 4ac , dove nella forma generale y = ax^2 + bx + c abbiamo a = 1 , b = -2(k-3) e c = -k + 15 .

Quindi, \Delta = (-2(k-3))^2 - 4(1)(-k+15) = 4(k^2 - 6k + 9) + 4k - 60 = 4k^2 - 24k + 36 + 4k - 60 = 4k^2 - 20k - 24 .

La parabola interseca l'asse delle x se \Delta > 0 :

4k^2 - 20k - 24 > 0 .

2. Per determinare quando la parabola interseca l'asse y in un punto di ordinata positiva, dobbiamo assicurarci che il termine noto -k + 15 sia positivo:

-k + 15 > 0 .

Ora risolviamo sia l'inequazione del discriminante che l'inequazione relativa al termine noto per trovare i valori di k che soddisfano entrambe le condizioni.

1. Per \Delta = 4k^2 - 20k - 24 > 0 :
4k^2 - 20k - 24 > 0 \implies k^2 - 5k - 6 > 0 \implies (k - 6)(k + 1) > 0.

Le soluzioni sono k oppure k > 6 .

2. Per
-k + 15 > 0 :
-k + 15 > 0 \implies k

Quindi, i valori di k che soddisfano entrambe le condizioni sono k \in (-\infty, -1) \cup (6, 15) .

\boxed{k \in (-\infty, -1) \cup (6, 15)} .

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