A jumping spider jumps from a log onto the ground below. Its height, h, in centimetres, as a function of time, t, in seconds, since it jumped can be modelled by the function h(t) = -490t^2 + 75t + 12.When does the spider land on the ground? ___ seconds (answer to 1 decimal place)

So I thought the answer would be 0.2 because parabolas are symmetric but my online assignment says it's not. Please help!

To determine when the spider lands on the ground, we need to find the time when the height, h(t), equals zero. In other words, we need to solve the equation -490t^2 + 75t + 12 = 0.

To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a = -490, b = 75, and c = 12.

Plugging in these values into the formula, we get:
t = (-75 ± √(75^2 - 4*(-490)*12)) / (2*(-490))

Simplifying this equation, we have:
t = (-75 ± √(5625 + 23520)) / (-980)
t = (-75 ± √29145) / (-980)

Now, we need to find the positive value of t since time cannot be negative. Let's solve for t:

t = (-75 + √29145) / (-980) ≈ 0.309 seconds (rounded to 3 decimal places)

Therefore, the spider lands on the ground approximately 0.309 seconds after it jumps.

To determine when the spider lands on the ground, we need to find the value of t when the height, h(t), is equal to zero. In other words, we want to find the value of t that satisfies the equation h(t) = 0.

Let's substitute h(t) = 0 into the given equation:

0 = -490t^2 + 75t + 12

We have a quadratic equation in standard form: at^2 + bt + c = 0, where a = -490, b = 75, and c = 12.

To solve for t, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Applying the values of a, b, and c into the formula, we have:

t = (-(75) ± √((75)^2 - 4(-490)(12))) / (2(-490))

Simplifying further:

t = (-75 ± √(5625 + 23520)) / (-980)
t = (-75 ± √29145) / (-980)

It seems that the value under the square root, 29145, is not a perfect square. To find the approximate decimal value of t, we can use a calculator.

Calculating √29145 results in approximately 170. L et's substitute that value back in:

t = (-75 ± 170) / (-980)

Now, we have two possible values for t:

t1 = (-75 + 170) / (-980) ≈ 0.102 seconds (rounded to 1 decimal place)
t2 = (-75 - 170) / (-980) ≈ -0.261 seconds (rounded to 1 decimal place)

Since time cannot be negative, we discard t2 since it doesn't make sense in the context of the problem. Therefore, the spider lands on the ground approximately 0.1 seconds after jumping.

it lands when the height is zero. That is when t ≈ 0.25

How did symmetry lead you to the answer of 0.2?