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course Mth 163
Question: `q007. Sketch a parabola through points (-4, 4), (1, -1) and (2, 4). The parabola should be symmetric about some vertical line.
Estimate the coordinates at which the parabola passes through the x axis.
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Your solution:
Previous Solution: The x coordinates are x=1.23 and x=-3.24
Those figures are pretty specific for an estimate.
It is expected that an estimate would be based on a hand-sketched graph.
The x axis numbers are x= 1.3 and x = -3.5
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Question: `q008. Solve the quadratic equation x^2 + 2 x - 4 = 0.
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Your solution:
Previous solution:
X^2 + 2x -4
(-1)^2 + 2*-1 -4
1-2-4
-5
The quadratic equation would be solved using either factoring or the quadratic formula.
-5 is not a solution to this equation.
You appear to have evaluated the equation for x = -1. Your result is correct, but that's not a solution to the equation.
X^2 + 2x -4 = 0 ; factor
(x+4) (x-1)
x+4=0; subtract 4 from both sides x-1=0; add 1 to both sides
x=-4 x=1
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This would be a good solution if your factoring was correct.
However this equation turns out to have an irrational solution, so you're not going to be able to factor it. This needs to be solved using the quadratic formula.
You can tell this isn't going to factor by evaluating the discriminant b^2 - 4 a c. You get 20, and the square root of 20 is not rational.
I really recommend that you solve this using the quadratic formula and submit it.
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Question: `q009. Verify that the points (-4, 4), (1, -1), (2, 4) lie on the graph of the parabola y = x^2 + 2 x - 4 = 0.
What is the value of y at the vertex of this parabola?
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Your solution:
Previous solution: The value of y is -5 at the vertex of this parabola. I got this by plugging in the x axis of the vertex and solving for y.
To show that a point lies on the graph you would plug the x coordinate of that point into the expression x^2 + 2 x - 4 and evaluate it. If the result agrees with the y coordinate of the point, then the point lies on the graph of the given parabola.
To get the y axis we have to plug in the x intercept from the point (2,4) into the equation x^2 + 2x -4 = 0.
Therefore we have:
(2)^2 + 2*2 -4 = 4 ; multiply
4 + 4 - 4 = 4 ; add and subtract
4=4
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You have verified that (2, 4) lies on the parabola.
You haven't found the y value at the vertex.
The vertex of the parabola y = a x^2 + b x + c has x coordinate -b / (2 a).
You should find the coordinates of the vertex, sketch these on your graph and be sure the coordinates make sense.
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Check my notes and my recommendations.
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