Question 1044241: I am not understanding these at all. Can someone help me with this example?
Given the quadratic function f(x) = 3x2 + 6x - 4
a. Find the vertex
b. Find the axis of symmetry
c. Find the x- and y-intercepts
d. State the domain and range of the function
e. Find the interval(s) over which the function is increasing
f. Find the interval(s) over which the function is decreasing
g. Graph the function
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! f(x) = 3x^2 +6x - 4
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This is a parabola that curves upward, here is its graph
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The general form of a polynomial for the given function is
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f(x) = ax^2 +bx +c, where a, b, c are real numbers
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The x coordinate for the vertex is given by
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x = -b /2a, for our given function we have
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x = -6 / (2 * 3) = -1
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substitute for x in the given function for the y coordinate of the vertex
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y = f(x) = 3(-1)^2 +6(-1) -4 = -7
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a) vertex is at (-1, -7)
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b) axis of symmetry is x = -1
half of the parabola is to the left and half of the parabola is to the right
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The y intercept is found by setting x = 0 and solve for y
y = 3(0)^2 +6(0) -4 = -4
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x intercept/s are found by solving the given function for x, when f(x) = 0
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3x^2 +6x -4 = 0
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use quadratic formula to solve for x
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x = (-6 + square root((-6)^2 -4*3*(-4))) / (2(3)) = (-3 + square root(21)) / 3
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x = (-6 - square root((-6)^2 -4*3*(-4))) / (2(3)) = (-3 - square root(21)) / 3
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c) y intercept is -4
x intercepts are (-3 + square root(21)) /3, (-3 - square root(21)) / 3
note the intercepts are where the graph of f(x) crosses the y axis(y intercept) and where f(x) crosses the x axis (x intercepts
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d) The domain and range of f(x) are the real numbers
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e) The graph of f(x) shows us the f(x) is increasing on the interval [-1, +infinity) and f(x) is decreasing on the interval [-infinity, -1]
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f) see graph above
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