**Question**

(i)Express in the form \(4x^{2}-12x\) in the form \(\left ( 2x+a \right )^{2}+b\)

**(ii)Hence ,or otherwise,find the set of values of x satisfying \(4x^{2}-12x> 7\)**

**▶️Answer/Explanation**

(i)To express \(4x^{2}-12x\) in the form \(\left(2x+a\right)^{2}+b\), we need to complete the square.

We start with \(4x^{2}-12x\). First, we factor out the common factor of 4:

\(4x^{2}-12x = 4(x^{2}-3x)\)

Next, we take half of the coefficient of the \(x\) term, which is \(-3/2\), square it, and add it inside the parentheses:

\(4\left [ \left ( x-\frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2} \right ]\)

\(4\left [ \left ( x-\frac{3}{2} \right )^{2}-\frac{9}{4} \right ]\)

\(4 \left ( x-\frac{3}{2} \right )^{2}-4\frac{9}{4}\)

\(\left [ 2\left ( x-\frac{3}{2} \right ) \right ]^{2}-9\)

\(\left [ 2\left ( x-\frac{3}{2} \right ) \right ]^{2}-9\)

\(\left ( 2x-3 \right )^{2}-9\)

So, \(4x^{2}-12x\) can be expressed in the form \(\left ( 2x-3 \right )^{2}-9\)

(ii) To solve the inequality \(4x^{2}-12x > 7\), we can first rewrite it as \(4x^{2}-12x-7 > 0\).

To factorize the quadratic expression, we can use the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

In this case, \(a = 4\), \(b = -12\), and \(c = -7\).

\(x = \frac{-(-12) \pm \sqrt{(-12)^{2}-4(4)(-7)}}{2(4)}\)

\(x = \frac{12 \pm \sqrt{144+112}}{8}\)

\(x = \frac{12 \pm \sqrt{256}}{8}\)

\(x = \frac{12 \pm 16}{8}\)

\(x_1 = \frac{12+16}{8} = \frac{28}{8} = \frac{7}{2}\)

\(x_2 = \frac{12-16}{8} = \frac{-4}{8} = -\frac{1}{2}\)

Therefore, the set of values of \(x\) satisfying \(4x^{2}-12x > 7\) is \(x > \frac{7}{2}\) or \(x < -\frac{1}{2}\).

**Question**

**Question**

(i) Express \( -x^{2}+6x-5\) in the form \(a(x+b)^{2}+c\), where a, b and c are constants.

The function \(f:x\rightarrow -x^{2}+6x-5\) defined for \(x\geq m\) ,where m is constant.

(ii) State the smallest value of m for which f is one-one.

(iii) For the case where m = 5, find an expression for \(f^{-1}(x)\) and state the domain of \(f^{-1}\).

**▶️Answer/Explanation**

(i) To express the quadratic function\( -x^{2}+6x-5\) in the form \(a(x+b)^{2}+c\) we need to complete the square.

First, let’s rewrite the quadratic function as follows:

\(-x^{2}+6x-5=-(x^{2}-6x+5)\)

Now, we want to complete the square inside the parentheses:

\(x^{2}-6x+5=(x^{2}-6x+9)-9+5=(x-3)^{2}-4\)

Substituting this back into the original expression:

\(-x^{2}+6x-5=-\left [ (x-3)^{2}-4 \right ]=-1\left ( x-3 \right )^{2}+4\)

So, the quadratic function can be expressed in the form \(a(x+b)^{2}+c\) as

\(-x^{2}+6x-5=-1(x-3)^{2}+4\)

Therefore, we have a=-1,b=3 and c=4.

(ii) For a quadratic function to be one-to-one (i.e., injective), it should pass the horizontal line test, which means each horizontal line intersects the graph at most once.

In this case, the quadratic function is f:x\rightarrow \(-x^{2}+6x-5\) for \(x\geq m\) , where m is a constant .

For the function to be one-to-one, the vertex of the parabola (which occurs at\( x=-\frac{b}{2a})\) should lie on or to the right of the line x=m.

Using the coefficients from the quadratic function, we have:

\(x=\frac{-6}{2(-1)}=3\)

Since the vertex is at x=3, we need m\leq 3 for the function to be one-to-one.

The smallest value of m that satisfies this condition is m=3.

(iii)\((x-3)^{2}=4-y\)

Correct order of operations

\(f^{-1}(x)=3+\sqrt{4-x}\)

Domain is \( x\leq 0\)

**Question**

(i) Express \(x^{2}+6x+2\) in the form \(\left ( x+a \right )^{2}+b\),where a and b are constants.

**(ii)Hence,or otherwise, find the set of values of x for which \(x^{2}+6x+2> 9\).**

**▶️Answer/Explanation**

(i) To express \(x^{2}+6x+2\) in the form \((x+a)^{2}+b\), we complete the square.

We start with \(x^{2}+6x+2\). To determine the value of \(a\), we take half of the coefficient of the \(x\) term, which is 6, and square it:

\(\left(\frac{6}{2}\right)^{2} = 9\)

\(x^{2}+6x+9+2-9 = (x+3)^{2}+(-7)\)

Therefore, \(x^{2}+6x+2\) can be expressed as \((x+3)^{2}-7\).

(ii) We have the inequality \(x^{2}+6x+2>9\). Let’s use the expression we derived in part (i) to simplify the inequality:

\((x+3)^{2}-7>9\)

\((x+3)^{2}>16\)

Taking the square root of both sides (considering both positive and negative roots):

\(x+3>4\) or \(x+3<-4\)

For \(x+3>4\): \(x>1\)

For \(x+3<-4\): \(x<-7\)

Therefore, the set of values of \(x\) that satisfies \(x^{2}+6x+2>9\) is \(x<-7\) or \(x>1\).

**Question**

**(i) Express \(x^{2}-2x-15\) in the form \((x+a)^{2}+b\).**

The function f is defined for p ≤ x ≤ q, where p and q are positive constants, by f : x → \(x^{2}\) − 2x − 15.

The range of f is given by c ≤ fx ≤ d, where c and d are constants.

(ii) State the smallest possible value of c.

For the case where c = 9 and d = 65,

(iii) find p and q,

(iv) find an expression for \(f^{-1}(x)\).

**▶️Answer/Explanation**

(i) To express \(x^{2}-2x-15\) in the form \((x+a)^{2}+b\), we complete the square. Let’s go through the steps:

\(x^{2}-2x-15 = (x^{2}-2x+1) – 16 = (x-1)^{2} – 16\)

Therefore, \(x^{2}-2x-15\) can be expressed as \((x-1)^{2} – 16\) in the desired form.

(ii) To determine the smallest possible value of \(c\), we need to find the minimum value of \(f(x)\) over its given domain. Since the coefficient of \(x^2\) is positive, the parabola opens upwards, and the minimum value occurs at the vertex.

The x-coordinate of the vertex can be found using the formula \(x = -\frac{b}{2a}\), where \(a\) is the coefficient of \(x^2\) and \(b\) is the coefficient of \(x\). In this case, \(a = 1\) and \(b = -2\). Therefore, the x-coordinate of the vertex is:

\(x = -\frac{-2}{2(1)} = 1\)

Substituting \(x = 1\) into the function \(f(x)\):

\(f(1) = 1^2 – 2(1) – 15 = 1 – 2 – 15 = -16\)

Thus, the minimum value of the function \(f\) is \(-16\). Therefore, the smallest possible value of \(c\) is \(-16\).

(iii) For the case where c=9 and d=65, we know that the range of is given by f \(c\leq f(x)\leq d\).Substituting the values, we have:

\(9\leq (x-1)^{2}-16\leq 65\)

\(25\leq (x-1)^{2}\leq 81\)

\(5\leq x-1\leq 9\)

\(6\leq x\leq 10\)

Therefore, the values of p and q are 6 and 10 respectively,

(iv) To find an expression for \(f^{-1}(x)\), we swap the roles of \(x\) and \(y\) and solve for \(y\) in terms of \(x\).

Starting from the original function \(f(x) = x^{2} – 2x – 15\):

\(x = y^{2} – 2y – 15\)

Rearranging the equation:

\(y^{2} – 2y = x + 15\)

Completing the square on the left side:

\(y^{2} – 2y + 1 = x + 15 + 1\)

\((y – 1)^{2} = x + 16\)

Taking the square root of both sides:

\(y – 1 = \pm \sqrt{x + 16}\)

Solving for \(y\):

\(y = 1 \pm \sqrt{x + 16}\)

So, the expression for \(f^{-1}=1 \pm \sqrt{x + 16}\)

**Question**

A curve for which \( \frac{\mathrm{d} y}{\mathrm{d} x}=7-x^{2}-6x \) pases tghrough the p[oint (3,-10)

(i)Find the equation of the curve.

(ii)Express \(7-x^{2}-6x\) in the form \(a-(x+b)^{2}\),where a and b are constants.

**(iii)Find the set of values of x for which the gradient of the curve is positive.**

**▶️Answer/Explanation**

(i) To find the equation of the curve, we need to integrate the given derivative with respect to \(x\). Integrating \( \frac{\mathrm{d} y}{\mathrm{d} x}=7-x^{2}-6x \) will give us the original function \(y\).

Integrating \(7-x^{2}-6x\) with respect to \(x\), we get:

\(y = \int (7-x^{2}-6x) \, dx = 7x – \frac{x^{3}}{3} – 3x^{2} + C\)

where \(C\) is the constant of integration.

(ii) To express \(7-x^{2}-6x\) in the form \(a-(x+b)^{2}\), we complete the square. Let’s start by rearranging the terms:

\(7-x^{2}-6x = -x^{2}-6x+7\)

Now, we want to find the value of \(b\) such that \(-x^{2}-6x+7\) can be written as \(-(x+b)^{2}\). We take half of the coefficient of the \(x\) term, which is \(-6\), and square it:

\(\left(\frac{-6}{2}\right)^{2} = 9\)

\(-x^{2}-6x+7 = -(x+3)^{2} + 9 + 7 = – (x+3)^{2} + 16\)

Therefore, \(7-x^{2}-6x\) can be expressed as \(- (x+3)^{2} + 16\).

(iii) The gradient of the curve is given by \(\frac{\mathrm{d} y}{\mathrm{d} x}\), which is \(7-x^{2}-6x\). To find the set of values of \(x\) for which the gradient is positive, we need to solve the inequality \(7-x^{2}-6x > 0\).

\(16-\left ( x+3 \right )^{2}>0\)

\(16-\left ( x^{2}+9+6x \right )>0\)

\(16-x^{2}-9-6x>0\)

\(-x^{2}-6x+7>0\)

\(x^{2}+6x-7<0\)

\((x-1)(x+7)<0\)

\(-7<x<1\)

Therefore, the set of values of \(x\) for which the gradient of the curve is positive is \(-7<x<1.\)

**Question**

**Question**

The graph of y = f(x) is transformed to the graph of \(y = 1 + f(\frac{1}{2}x)\).

Describe fully the two single transformations which have been combined to give the resulting transformation.

**▶️Answer/Explanation**

The given transformation \(y = 1 + f\left(\frac{1}{2}x\right)\) involves two single transformations:

Horizontal compression by a factor of 2: This transformation is represented by the expression \(\frac{1}{2}x\). It indicates that the x-values of the original function \(f(x)\) have been halved, resulting in a horizontal compression. This means that the graph is narrower compared to the original graph. Points on the graph are shifted closer to the y-axis.

Vertical shift upward by 1 unit: This transformation is represented by the constant term \(1\) added to the function \(f\left(\frac{1}{2}x\right)\). It indicates that the entire graph has been shifted vertically by 1 unit in the positive y-direction. The effect of this transformation is that all the points on the graph are moved upward by 1 unit.

In summary, the original graph of \(y = f(x)\) is first horizontally compressed by a factor of 2, making it narrower, and then shifted vertically upward by 1 uni

**Question**

**Question**

The function f is defined by \(f(x)=-2x^{2}+12x-3\) for \(\varepsilon R\)

(i) Express \(-2x^{2}+12x-3 \)in the from \(2(x+a)^{a}+b,\)where a and b are constants.

(ii) State the greatest value of f(x).

The function g is defined by g(x) = 2x + 5 .

(iii) Find the values of x for which gf(x )+1=0.

**▶️Answer/Explanation**

(i) To express the function \(-2x^2 + 12x – 3\) in the form \(2(x + a)^2 + b\), we need to complete the square. Let’s go through the steps:

\(-2x^2 + 12x – 3 = -2(x^2 – 6x) – 3 = -2(x^2 – 6x + 9 – 9) – 3\)

Expanding and simplifying:

\(-2(x^2 – 6x + 9) + 18 – 3 = -2(x – 3)^2 + 15\)

Therefore, \(-2x^2 + 12x – 3\) can be expressed as \(2(x – 3)^2 + 15\) in the desired form, where \(a = 3\) and \(b = 15\).

(ii) To find the greatest value of \(f(x)\), we observe that the quadratic term \(-2x^2\) has a negative coefficient. This means the graph of \(f(x)\) opens downward and the vertex represents the maximum point. The x-coordinate of the vertex can be found using the formula \(-\frac{b}{2a}\) for a quadratic in the form \(ax^2 + bx + c\).

For \(f(x) = -2x^2 + 12x – 3\), the coefficient of the quadratic term is \(a = -2\) and the coefficient of the linear term is \(b = 12\). Using the formula, we can find:

\(x_{\text{vertex}} = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3\)

Substituting this x-coordinate into the function, we find:

\(f(3) = -2(3)^2 + 12(3) – 3 = -18 + 36 – 3 = 15\)

Therefore, the greatest value of \(f(x)\) is 15.

(iii) To find the values of \(x\) for which \(g(f(x)) + 1\), we need to substitute \(f(x)\) into \(g(x)\) and solve for \(x\):

\(g(f(x)) + 1 = 2f(x) + 5 + 1 = 2(-2x^2 + 12x – 3) + 6 = -4x^2 + 24x – 1+1\)

Now, we set this expression equal to \(x\) and solve for \(x\):

\(-4x^2 + 24x = 0\)

\(4x^{2}-24x=0\)

\(4x(x-6)=0\)

\(x=6\) and \(x=0\)