The parabola with the general equation y= ax^2 + bx+ 9 where0 < a < 10 and 0 < b < 20 touches the x-axis at one point only. The graph passes through the point (1, 25). Find the values of a and b.
Please help.
Have you thought about narrowing it down a bit?
Suggestions...
1) sketch it with the information given (one x intercept)
2) sub in your point on the quadratic and see what results :)
What do those two suggestions help you to see?
For the equation ax^2 + bx+ 9 = 0 to have only one solution, (the parabola touches the x-axis at one point only) , the discriminant has to be zero, that is
b^2 - 4a(9) = 0
b^2 = 36a
a = b^2/36
Also MsPi suggested subbing in the given point (1,25)
25 = a(1)^2 + b(1) + 9
a+b = 16
then b^2/36 + b = 16
b^2 + 36b - 576 = 0
b-12)(b+48) = 0
b = 12 or b = -48
but you said 0<b<20
so b = 12
ok, you finish it.
To find the values of a and b, we can start by using the fact that the parabola touches the x-axis at one point only.
When a parabola touches the x-axis at one point, it means that the discriminant (b^2 - 4ac) of the quadratic equation is equal to 0.
In this case, the quadratic equation for the given parabola is:
y = ax^2 + bx + 9
Let's substitute y=0 (since the parabola touches the x-axis) into the equation and find the discriminant:
0 = ax^2 + bx + 9
0 = a(1)^2 + b(1) + 9
0 = a + b + 9
The discriminant is given by:
D = b^2 - 4ac
Since the discriminant is 0, we can substitute the values of a and b from the earlier equation:
0 = b^2 - 4(a)(9)
Simplifying further, we get:
0 = b^2 - 36a
Since we know that a is between 0 and 10, and b is between 0 and 20, we have two possibilities:
1. b^2 - 36a = 0
2. b^2 - 36a > 0
Let's solve equation 1:
b^2 - 36a = 0
b^2 = 36a
Since a can't be 0, we can divide both sides by a:
b^2/a = 36
b^2 = 36a
Let's solve equation 2:
b^2 - 36a > 0
b^2 > 36a
Now, we know that the parabola passes through the point (1, 25).
Let's substitute these values into the equation:
25 = a(1)^2 + b(1) + 9
25 = a + b + 9
a + b = 16
We can solve this equation simultaneously with equation 1 or equation 2 to find the values of a and b.
Let's solve it with equation 1:
b^2 = 36a
a + b = 16
From the second equation, we can rearrange to get b = 16 - a.
Substituting this into the first equation:
(16 - a)^2 = 36a
Expanding and simplifying:
256 - 32a + a^2 = 36a
Rearranging:
a^2 + 36a - 32a + 256 = 0
a^2 + 4a + 256 = 0
This quadratic equation can be factored as:
(a + 4)(a + 64) = 0
The solutions are:
a + 4 = 0
a = -4
and
a + 64 = 0
a = -64
However, since a cannot be less than 0, the only valid solution is a = -4.
Substituting this into equation 2:
b^2 > 36a
b^2 > 36(-4)
b^2 > -144
Since the square of any real number is always positive, we can conclude that b^2 > 0.
Therefore, b must be greater than 0.
In conclusion, the values of a and b are:
a = -4
b > 0
To find the values of a and b, we can use the information provided in the question.
Firstly, we know that the parabola touches the x-axis at one point only. This means that the parabola has a single root, or in other words, its discriminant is equal to 0.
The discriminant of a quadratic equation of the form ax^2 + bx + c = 0 is given by Δ = b^2 - 4ac. In our case, since the parabola touches the x-axis at one point only, Δ = 0.
Substituting the given values into the discriminant equation, we have:
0 = b^2 - 4(a)(9)
Expanding, we get:
0 = b^2 - 36a
Next, we know that the graph passes through the point (1, 25). Substituting these values into the equation of the parabola, we have:
25 = a(1)^2 + b(1) + 9
Simplifying, we get:
25 = a + b + 9
Rearranging, we have:
a + b = 16 ---(1)
Now, we have two equations:
1) 0 = b^2 - 36a
2) a + b = 16
We can solve these equations simultaneously to find the values of a and b.
From equation (1), we have a = 16 - b. Substituting this into equation (2), we get:
16 - b + b = 16
Simplifying, we have:
16 = 16
This equation is true, which means that there are infinite possible values for a and b that satisfy these conditions.
Therefore, we cannot determine the exact values of a and b based on the given information.