Sketch the graph of $y = 7^x$, $x
eq ext{R}$, showing the coordinates of any points at which the graph crosses the axes - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 4
Question 2
Sketch the graph of $y = 7^x$, $x
eq ext{R}$, showing the coordinates of any points at which the graph crosses the axes.
Solve the equation $7^{2x} - 4(7^x) + ... show full transcript
Worked Solution & Example Answer:Sketch the graph of $y = 7^x$, $x
eq ext{R}$, showing the coordinates of any points at which the graph crosses the axes - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 4
Step 1
Sketch the graph of $y = 7^x$
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Answer
To sketch the graph of the function y=7x, we need to analyze its behavior:
Identify Axes Intercepts:
The graph crosses the y-axis when x=0:
y=70=1
Therefore, the point is (0,1).
The graph does not cross the x-axis, as the function never reaches zero for any real x.
General Behavior:
As x approaches negative infinity, y approaches 0 but never touches it.
As x increases, y increases rapidly because it is an exponential function.
Sketch the Graph:
The graph starts near the x-axis and passes through the point (0,1), rising steeply as x increases.
Step 2
Solve the equation $7^{2x} - 4(7^x) + 3 = 0$
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Answer
To solve the equation, we can use substitution:
Substitution: Let u=7x. Then, the equation becomes:
u2−4u+3=0
Factor the Quadratic: This can be factored as:
(u−3)(u−1)=0
Solve for u:
Setting each factor to zero gives:
u−3=0ightarrowu=3
u−1=0ightarrowu=1
Back Substitute for x:
For u=3:
ightarrow x = rac{ ext{log}(3)}{ ext{log}(7)} \ ext{Calculate: } x \ ≈ 0.564.$$
For u=1:
ightarrow x = 0 \ ext{Exact value: } x = 0.$$
Final Answers:
Therefore, the solutions to the original equation are:
xext(to2decimalplaces):x=0.56extandx=0.