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State the x-intercepts of the graph of the equation. Then find the coordinates of the vertex. The vertex and an ordered pair must be graphed on both sides of the vertex.

4. f(x) = 2x^2 – 9x + 9

User Rsteckly
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1 Answer

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Answer:

the point (5/2, -27/2) is also on the graph of the function. By plotting these points, we can sketch the graph of the function.

Explanation:

To find the x-intercepts of the graph of f(x) = 2x^2 – 9x + 9, we set f(x) equal to zero and solve for x:

2x^2 – 9x + 9 = 0

We can use the quadratic formula to solve for x:

x = [-(b) ± sqrt((b^2) - 4(a)(c))] / 2(a)

Plugging in the values from our equation, we get:

x = [-( -9 ) ± sqrt( (-9)^2 - 4(2)(9) )] / 2(2)

x = [9 ± sqrt(9)] / 4

x = (9 ± 3) / 4

x = 3/2, 3

Therefore, the x-intercepts of the graph are (3/2, 0) and (3, 0).

To find the coordinates of the vertex, we use the formula:

x = -b / 2a

We know that a = 2 and b = -9 from our equation, so we get:

x = -(-9) / 2(2)

x = 9/4

To find the corresponding y-coordinate of the vertex, we plug this value of x back into our equation:

f(9/4) = 2(9/4)^2 – 9(9/4) + 9

f(9/4) = 81/8 - 81/4 + 9

f(9/4) = -27/8

Therefore, the vertex of the graph is (9/4, -27/8).

To graph this function, we can plot the vertex at (9/4, -27/8) and the x-intercepts at (3/2, 0) and (3, 0). We can also plot a point on either side of the vertex by choosing an x-value and finding the corresponding y-value using the equation. For example, if we plug in x = 2, we get:

f(2) = 2(2)^2 - 9(2) + 9

f(2) = -7

So the point (2, -7) is on the graph of the function. Similarly, if we plug in x = 5/2, we get:

f(5/2) = 2(5/2)^2 - 9(5/2) + 9

f(5/2) = -27/2

So the point (5/2, -27/2) is also on the graph of the function. By plotting these points, we can sketch the graph of the function.

User Per Arneng
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