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A parabola k(x) = ax2 and a line p(x)=mx+c are shown in the diagram. The vertex of k is at the origin. Point C(0;-6) is on line p. The gradient of AD is 1. Point B is equidistant from A to C. Find the

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6 parts
4.1

y≤0y \le 0

4.2

x∈Rx \in \mathbb{R}

4.3

x - 6

4.4

−13x2-\frac{1}{3}x^2

4.5
(-6; -12)
4.7

45∘45^\circ

Step-by-step solution

15 steps

Here are the solutions to Question 4.

Part 4.1

The graph kk is a parabola with the equation k(x)=ax2k(x) = ax^2. From the diagram, the parabola opens downwards, which means a<0a < 0. The vertex of this parabola is at the origin (0,0)(0,0), which is its maximum point.
The range of kk is all yy-values less than or equal to the maximum value.

Range of k:y≤0\text{Range of } k: \boxed{y \le 0}

Part 4.2

x∈Rx \in \mathbb{R}

The graph pp is a straight line. The domain of any linear function is all real numbers.

Domain of p:x∈R\text{Domain of } p: \boxed{x \in \mathbb{R}}

Part 4.3x - 6

Given that the gradient of AD (which is line pp) is 1, and its equation is p(x)=mx+cp(x) = mx + c.

  1. Substitute the given gradient m=1m=1 into the equation.

    p(x)=1x+cp(x) = 1x + c p(x)=x+cp(x) = x + c

  2. Use point C(0;−6)(0;-6), which lies on line pp, to find the value of cc.

    Substitute x=0x=0 and p(x)=−6p(x)=-6:

    −6=0+c-6 = 0 + c c=−6c = -6

  3. Write the equation of AD.

    p(x)=x−6p(x) = \boxed{x - 6}

Part 4.4

−13x2-\frac{1}{3}x^2

Determine the equation of kk in the form of k(x)=ax2k(x) = ax^2 if B is equidistant from A to C.

  1. Find the coordinates of point B.

    Since B is equidistant from A(6;0)(6;0) and C(0;−6)(0;-6), B is the midpoint of AC.
    The midpoint formula is (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right).

    B=(6+02,0+(−6)2)B = \left(\frac{6+0}{2}, \frac{0+(-6)}{2}\right) B=(62,−62)B = \left(\frac{6}{2}, \frac{-6}{2}\right) B=(3,−3)B = (3, -3)

  2. Use the coordinates of B to find the value of aa in k(x)=ax2k(x) = ax^2.

    Point B(3;−3)(3;-3) lies on the parabola k(x)=ax2k(x) = ax^2.
    Substitute x=3x=3 and k(x)=−3k(x)=-3:

    −3=a(3)2-3 = a(3)^2 −3=9a-3 = 9a a=−39a = \frac{-3}{9} a=−13a = -\frac{1}{3}

  3. Write the equation of kk.

    k(x)=−13x2k(x) = \boxed{-\frac{1}{3}x^2}

Part 4.5(-6; -12)

Calculate the co-ordinates of D.

  1. Set the equations of line pp and parabola kk equal to each other to find their intersection points.

    We have p(x)=x−6p(x) = x - 6 and k(x)=−13x2k(x) = -\frac{1}{3}x^2.

    x−6=−13x2x - 6 = -\frac{1}{3}x^2

  2. Solve the quadratic equation for xx.

    Multiply by 3 to clear the fraction:

    3(x−6)=−x23(x - 6) = -x^2 3x−18=−x23x - 18 = -x^2

    Rearrange into standard form:

    x2+3x−18=0x^2 + 3x - 18 = 0

    Factor the quadratic equation:

    (x+6)(x−3)=0(x + 6)(x - 3) = 0

    The solutions for xx are x=−6x = -6 or x=3x = 3.

  3. Identify the coordinates of D.

    We know that x=3x=3 corresponds to point B(3;−3)(3;-3). Therefore, x=−6x=-6 must correspond to point D.
    Substitute x=−6x=-6 into the equation of line pp to find the yy-coordinate of D:

    y=x−6y = x - 6 y=−6−6y = -6 - 6 y=−12y = -12

    The coordinates of D are (−6;−12)\boxed{(-6; -12)}.

Part 4.6· 3 steps

What kind of a triangle is △ACO\triangle ACO according to the lengths of the sides and give a reason.

  1. Identify the coordinates of the vertices.

    A(6;0)(6;0), C(0;−6)(0;-6), O(0;0)(0;0).

  2. Calculate the lengths of the sides AO and CO.

    Length of AO: The distance from (0,0)(0,0) to (6,0)(6,0) is 6 units.
    Length of CO: The distance from (0,0)(0,0) to (0,−6)(0,-6) is 6 units.

  3. Classify the triangle based on side lengths.

    Since AO=CO=6AO = CO = 6 units, two sides of the triangle are equal in length.
    Therefore, △ACO\triangle ACO is an isosceles triangle.
    Reason: Two sides are equal in length (AO=COAO = CO).

Part 4.7

45∘45^\circ

Without any calculations give the value of C^\hat{C}.

  1. Identify the vertices and their positions relative to the origin.

    A(6;0)(6;0) is on the positive x-axis.
    C(0;−6)(0;-6) is on the negative y-axis.
    O(0;0)(0;0) is the origin.

  2. Determine the angle at O.

    Since AO lies on the x-axis and CO lies on the y-axis, the angle ∠AOC\angle AOC is a right angle, i.e., 90∘90^\circ.

  3. Use properties of △ACO\triangle ACO.

    From Question 4.6, we know that AO=CO=6AO = CO = 6 units.
    Therefore, △ACO\triangle ACO is an isosceles right-angled triangle (right angle at O).
    In an isosceles right-angled triangle, the two non-right angles are equal and each measures 45∘45^\circ.
    Thus, C^=∠ACO=45∘\hat{C} = \angle ACO = 45^\circ.

    C^=45∘\hat{C} = \boxed{45^\circ}
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