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Find the vertex and x-intercept(s) for the quadratic function f(x)=4.9x^2-28x+40

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Answer:

To find the vertex and x-intercepts for the quadratic function f(x)=4.9x^2-28x+40:

First, we can find the x-coordinate of the vertex using the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic function.

In this case, a = 4.9 and b = -28, so:

x = -(-28)/2(4.9) = 2.86

Next, we can find the y-coordinate of the vertex by plugging in this x value into the original function:

f(2.86) = 4.9(2.86)^2 - 28(2.86) + 40 = 6.65

So the vertex is at (2.86, 6.65).

To find the x-intercepts, we can set the function equal to zero and solve for x:

4.9x^2 - 28x + 40 = 0

We can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Plugging in the coefficients, we get:

x = (28 ± sqrt(28^2 - 4(4.9)(40))) / 2(4.9)

Simplifying:

x = (28 ± sqrt(144.4)) / 9.8

x = 5.36 or 1.43

So the x-intercepts are (5.36, 0) and (1.43, 0).

User Keshavram Kuduwa
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