Answer:
-- x intercept
-- y intercept
Step-by-step explanation:
Given
Required
Determine the x and y intercept
For x intercept.
Set
So, we have:
Collect Like Terms
Make x the subject
Hence, the x intercept is:
For the y intercept;
Set
So, we have:
Hence, the y intercept is:
We are given :
Step 1: factor the part inside square root
The function given inside square root is of quadratic form.
So let us try to factorise it using AC method.
Here A*C = 4*25 = 100
so we have to find factors of 100 that add up to give -20.
the two factors are -10 and -10.
Rewriting the function :
=
=
=
Step 2:
Now we take square root of the factorised form
=
Answer : (2x-5)
Problem 1
Since point P is the tangent point, this means angle OPT is a right angle
angle OPT = 90 degrees
Let's use the Pythagorean theorem to find the missing side 'a'
a^2 + b^2 = c^2
a^2 + 7^2 = 12^2
a^2 + 49 = 144
a^2 = 144 - 49
a^2 = 95
a = sqrt(95)
a = 9.7467943
a = 9.7
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Let's use the sine and arcsine rule to find angle y
sin(angle) = opposite/hypotenuse
sin(y) = 7/12
y = arcsin( 7/12 )
y = 35.6853347
y = 35.7
The value of y is approximate. Make sure your calculator is in degree mode. Arcsine is the same as inverse sine or
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Problem 2
Focus on triangle OHP. This may or may not be a right triangle. The goal is to test if it is or not.
We use the converse of the Pythagorean theorem to check.
Recall that the converse of the Pythagorean theorem says: If a^2+b^2 = c^2 is a true equation, then the triangle is a right triangle. The value of c is always the longest side. The order of a and b doesn't matter.
In this case we have
Which leads to
a^2 + b^2 = c^2
7^2 + 13^2 = 16^2
49 + 169 = 256
218 = 256
We get a false equation at the end, since both sides aren't the same number, which means the original equation is false.
We don't get a^2 + b^2 = c^2 to be true, therefore this triangle is not a right triangle.
Consequently, this means angle OPH cannot possible be 90 degrees (if it was then we'd have a right triangle). Therefore, point P is not a tangent point.
You follow the same basic idea for triangle OQH and show that point Q is not a tangent point either. Or you could use a symmetry argument to note that triangle OPH is a mirror reflection of triangle OQH over the line segment OH. This implies that whatever properties triangle OPH has, then triangle OQH has them as well for the corresponding pieces.