Detailed Answer

a)

To find the y-intercept, set \( x = 0 \):

\[ f(0) = (2 \cdot 0 - 1)(0 + 2) \]

\[ f(0) = (-1)(2) \]

\[ f(0) = -2 \]

Therefore, the y-intercept is \( (-2, 0) \).

b)

To find the x-intercepts, set \( f(x) = 0 \):

\[ (2x - 1)(x + 2) = 0 \]

Solve for \( x \):

\[ 2x - 1 = 0 \]

\[ 2x = 1 \]

\[ x = \frac{1}{2} \]

One x-intercept is \( \left( \frac{1}{2}, 0 \right) \).

Next,

\[ x + 2 = 0 \]

\[ x = -2 \]

Therefore, the x-intercepts are \( \left( \frac{1}{2}, 0 \right) \) and \( (-2, 0) \).

c)

The expanded form of this function is: \( 2x^2 + 3x - 2 \), so using the vertex formula: \( x = -\frac{b}{2a} \), we get:

\[ x = -\frac{3}{2 \cdot 2} \]

\[ x = -\frac{3}{4} \]

To find \( f\left( -\frac{3}{4} \right) \):

\[ f\left( -\frac{3}{4} \right) = (2 \cdot \left( -\frac{3}{4} \right) - 1)(\left( -\frac{3}{4} \right) + 2) \]

\[ f\left( -\frac{3}{4} \right) = (-\frac{3}{2} - 1)(\frac{5}{4}) \]

\[ f\left( -\frac{3}{4} \right) = (-\frac{5}{2})(\frac{5}{4}) \]

\[ f\left( -\frac{3}{4} \right) = -\frac{25}{8} \]

Thus, the coordinates of the vertex are \( \left( -\frac{3}{4}, -\frac{25}{8} \right) \).

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Detailed Answer

To have two distinct real roots, the discriminant of the quadratic equation must be greater than zero:

\[ \Delta = b^2 - 4ac > 0 \]

For \( x^2 + kx + 3 = 0 \), where \( a = 1 \), \( b = k \), and \( c = 3 \), the discriminant is:

\[ k^2 - 12 > 0 \]

Solving for \( k \), we get:

\[ k^2 > 12 \]

\[ |k| > \sqrt{12} \]

\[ |k| > 2\sqrt{3} \]

Thus, \( k \) must satisfy \( k > 2\sqrt{3} \) or \( k < -2\sqrt{3} \).

Therefore, \( k \) must be greater than \( 2\sqrt{3} \) or less than \( -2\sqrt{3} \) to ensure the equation has two distinct real roots.

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Detailed Answer

a) We can see that one zero is at \( x = -2 \) and the second one is at \( x = 4 \). Since we know that \( a,b \geq 0 \) it has to be that \( a = 2 \) and \( b = 4 \), resulting in the final equation:

\[ f(x) = (x+2)(x-4) \]

b) Knowing that the vertex is at the point \( (1,-9) \), we know that the equation has to be:

\[ f(x) = (x-1)^2 - 9 \]

Therefore, \( h=1, k=-9 \).

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Detailed Answer

a) (i) y-intercept:

\[ y = 0^2 + 5 * 0 + 4 = 4 \]

Therefore, the y-intercept is \( (0, 4) \).

a) (ii)

To find the x-intercepts, solve \( x^2 + 5x + 4 = 0 \) using the quadratic formula:

\[ x = \frac{-5 \pm \sqrt{5^2 - 4 * 1 \cdot 4}}{2 * 1} \]

\[ x_1 = \frac{-5 + 3}{2} = -1 \]

\[ x_2 = \frac{-5 - 3}{2} = -4 \]

Therefore, the x-intercepts are \( (-1, 0) \) and \( (-4, 0) \).

b)

Using the answer from part (a) we have that:

\[ y = (x + 1)(x + 4) \]

c) Using the formula: \( x = -\frac{b}{2a} \)

\[ x = -\frac{5}{2} = -2.5 \]

\[ y = (-2.5)^2 + 5 * + 4 \]

\[ y = 6.25 - 12.5 + 4 \]

\[ y = -2.25 \]

Therefore, the coordinates of the vertex are \( (-2.5, -2.25) \).

d)

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Detailed Answer

a) Knowing that the trees surrounding the park have to cover the length of 200m, we have that:

\[ 200 = 2x + 2y \implies 100 = x + y \implies y = 100 - x \]

The formula for the area of the park is:

\[ A = x * y \]

By substituting \( y = 100 - x \):

\[ A = x * (100-x) = 100x - x^2 \]

b) We can plot it in the GDC and find the maximum area of \( 2500m^2 \).

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Detailed Answer

a)

Given the distance between the x-intercepts \( A \) and \( B \) is 10 units, the x-intercepts are symmetrically located around the vertex at \( x = 6 \). Therefore, we can write:

\[ A = (6 - 5, 0) = (1, 0) \]

\[ B = (6 + 5, 0) = (11, 0) \]

b) Based on the answer from part (a) we know that:

\[ y = (x - 1)(x-11)\]

\[ y = x^2 - 12x + 11 \]

So we know that:

\[ k = -12 \]

\[ t = 11 \]

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Detailed Answer

We know that for two roots to be equal, the discriminant has to be equal to 0, so:

\[ (a-2)^2 - 4*2* (-\frac{9}{8}a) = 0 \]

\[ (a^2 - 4a + 4) + 9a = 0 \]

\[ a^2 + 5a + 4 = 0 \]

\[ (a+4)(a+1) = 0 \]

\[ a = -4, \ a = -1\]

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Detailed Answer

a) (i)

The points of intersection are \( (1, c) \) and \( (5, c) \), so the axis of symmetry is the midpoint of these x-coordinates:

\[ x = \frac{1 + 5}{2} = 3 \]

Therefore, the axis of symmetry is \( x = 3 \).

a) (ii)

Since the axis of symmetry is \( x = 3 \), we can equate this to the formula:

\[ -\frac{12}{2a} = 3 \]

Solving for \( a \), we get:

\[ -\frac{6}{a} = 3 \]

\[ -6 = 3a \]

\[ a = -2 \]

b)

Let's substitute \( a = -2 \) and \( d = 5 \) into the function:

\[ y = -2x^2 + 12x + 5 \]

Since the line \( L \) intersects the graph at points \( (1, c) \) and \( (5, c) \), we can substitute \( x = 1 \) to find \( c \):

\[ c = -2(1)^2 + 12(1) + 5 \]

\[ c = -2 + 12 + 5 \]

\[ c = 15 \]

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Detailed Answer

\[ \log_p(-3x^2 - \sqrt{48}x) = 2 \]

\[ -3x^2 - \sqrt{48}x = p^2 \]

\[ -3x^2 - \sqrt{48}x - p^2 = 0 \]

For it to have only one solution, the discriminant has to be equal to zero:

\[ \Delta = (-\sqrt{48})^2 - 4(-3)(-p^2) \]

\[ \Delta = 48 - 12p^2 \]

\[ 48 - 12p^2 = 0 \]

\[ 12p^2 = 48 \]

\[ p^2 = 4 \]

Since we know that \( p > 0 \):

\[ p = 2 \]

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Detailed Answer

a) Since the function \( f(x) \) has zeros at \( -2 \) and \( 7 \), we have \( p = -2 \) and \( q = 7 \).

\[ f(x) = a(x + 2)(x - 7) \]

b) We know that the function passes through point \( Q(6, 4) \):

\[ 4 = a(6 + 2)(6 - 7) \]

\[ 4 = a(8)(-1) \]

\[ 4 = -8a \]

\[ a = -\frac{1}{2} \]

c) The axis of symmetry is the midpoint of the roots \( -2 \) and \( 7 \):

\[ \text{Axis of symmetry} = \frac{-2 + 7}{2} = \frac{5}{2} = 2.5 \]

d) The vertex lies on the axis of symmetry, so we plug \( x = 2.5 \) into \( f(x) \) to find the y-coordinate:

\[ f(2.5) = -\frac{1}{2}(2.5 + 2)(2.5 - 7) \]

\[ f(2.5) = -\frac{1}{2}(4.5)(-4.5) \]

\[ f(2.5) = -\frac{1}{2}(20.25) \]

\[ f(2.5) = 10.125 \]

The vertex is at \( (2.5, 10.125) \).

e) To find the slope of the tangent at \( R(4.5, 5) \), we first differentiate \( f(x) \):

\[ f(x) = -\frac{1}{2}(x + 2)(x - 7) \]

\[ f(x) = -\frac{1}{2}(x^2 - 5x - 14) \]

\[ f(x) = -\frac{1}{2}x^2 + \frac{5}{2}x + 7 \]

\[ f'(x) = -x + \frac{5}{2} \]

The slope of the tangent at \( x = 4.5 \):

\[ f'(4.5) = -4.5 + \frac{5}{2} = -4.5 + 2.5 = -2 \]

Using the point-slope form of the line equation with point \( R(4.5, 5) \): \[ y - 5 = -2(x - 4.5) \]

\[ y - 5 = -2x + 9 \]

So, we know that the equation of the tangent line is:

\[ y = -2x + 14 \]

Plugging the point \( x = 7 \): \[ y = -2(7) + 14 \]

\[ y = -14 + 14 = 0 \]

Thus, the line \( L_1 \) has a y-intercept at \( x = 7 \).

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