I need help with pre-cal?

Question

1: List all possible rational roots for the equation: 3x^4-5x^2+25=0

2: Solve for x: x^19 − 2x^18 + 7x^17 − 12x^6 + 6x^15 = 0

3: Given a vertex (h, k) and a point (x, y) on the parabola, derive the quadratic function: (h, k) = (2, 3) and (x, y) = (5, 12).

4: For the polynomial function ƒ(x) = −2x4 + 8x2 − 8, find all local and global extrema.

5: Given a vertex (h, k) and a point (x, y) on the parabola, derive the quadratic function: (h, k) = (−5, 3) and (x, y) = (2, 9).

please help me, i'm so tired and I really need help :( please.

Answer 1

1).

The polynomial is .

Rewrite the polynomial as .

Compare the above equation with standard form of the quadratic equation .

.

Solution of the quadratic equation .

Here, the given equation is a fourth degree polynomial, so

.

The possible rational roots for the given equation are  .

Answer 2

3).

The vertex is (2, 3) and a point is (5, 12).

The standard form of the equation of a parabola with vertex (h, k ) and axis of symmetry x = h is y = a(x - h )^2 + k.

Substitute the values of (h, k ) = (2, 3) and (x, y ) = (5, 12) in y = a(x - h )^2 + k.

12 = a( 5 - 2)^2 + 3

12 = a(3)^2 + 3

12 = 9a + 3

9a = 12 - 3 = 9

⇒ a = 1.

Substitute the values of a = 1 and (h, k ) = (2, 3) in y = a(x - h )^2 + k.

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

y = x^2 - 4x + 4 + 3

y = x^2 - 4x + 7.

The quadratic function is y = x^2 - 4x + 7.

5).

The vertex is (- 5, 3) and a point is (2, 9).

The standard form of the equation of a parabola with vertex (h, k ) and axis of symmetry x = h is y = a(x - h )^2 + k.

Substitute the values of (h, k ) = (- 5, 3) and (x, y ) = (2, 9) in y = a(x - h )^2 + k.

9 = a( 2 + 5)^2 + 3

9 = a(7)^2 + 3

9 = 49a + 3

49a = 9 - 3 = 6

⇒ a = 6/49.

Substitute the values of a = 6/49 and (h, k ) = (- 5, 3) in y = a(x - h )^2 + k.

y = (6/49)(x + 5 )^2 + 3

y = (6/49)(x^2 + 10x + 25) + 3

y = (6/49)x^2 + (60/49)x + (297/49).

The quadratic function is y = (6/49)x^2 + (60/49)x + (297/49).

Answer 3

4).

The polynomial function is f( x ) = - 2x^4 + 8x^2 - 8.

Apply first derivative with respect to x.

f '( x ) = - 8x^3 + 16x

Apply second derivative with respect to x.

f ''( x ) = - 24x^2 + 16

To find the critical values, or does not exist.

f '( x ) = - 8x^3 + 16x = 0

x^3 - 2x = 0

x(x^2 + 2) = 0

⇒ x = 0 and x = ±√2.

Relative extrema :

f ''(0) = - 24(0)^2 + 16 = 16 > 0 hence (0,f(0)) = ( 0, - 8 ) is a relative minimum.

f ''( ± √2 ) = - 24(±√2)^2 + 16 = - 48 + 16 = - 32 <  0 hence ( ± √2, 0 ) are the global maxima.