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
(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 \(x>1\).