Question

Use any method to locate the zeros of y=−3x2+12x+15.(1 point)
Responses

(0,15)
left parenthesis 0 comma 15 right parenthesis

(1,0)(−5,0)
left parenthesis 1 comma 0 right parenthesis left parenthesis negative 5 comma 0 right parenthesis

There are no zeros.
There are no zeros.

(−1,0)(5,0)
left parenthesis negative 1 comma 0 right parenthesis left parenthesis 5 comma 0 right parenthesis

Answer 1

To find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we can set the equation to zero and solve for \( x \):

\[
-3x^2 + 12x + 15 = 0
\]

First, let's simplify the equation by multiplying through by -1 to make the leading coefficient positive:

\[
3x^2 - 12x - 15 = 0
\]

Next, we can use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, \( a = 3 \), \( b = -12 \), and \( c = -15 \). Plugging these values into the formula:

\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(-15)}}{2(3)}
\]

Calculating step by step:

1. \( -(-12) = 12 \)
2. \( (-12)^2 = 144 \)
3. \( 4(3)(-15) = -180 \)
4. Hence, \( b^2 - 4ac = 144 + 180 = 324 \)

Now, substituting these into the formula:

\[
x = \frac{12 \pm \sqrt{324}}{6}
\]

Calculate \( \sqrt{324} \):

\[
\sqrt{324} = 18
\]

Now substituting back into the equation:

\[
x = \frac{12 \pm 18}{6}
\]

This gives us two solutions:

1. \( x = \frac{12 + 18}{6} = \frac{30}{6} = 5 \)
2. \( x = \frac{12 - 18}{6} = \frac{-6}{6} = -1 \)

Thus, the zeros of the function are \( x = 5 \) and \( x = -1 \).

Hence, the coordinates of the zeros are:

\[
(-1, 0) \text{ and } (5, 0)
\]

So the correct response is:

\[
\text{(−1,0)(5,0) }
\]

Which reads as \( \left( -1, 0 \right) \quad \left( 5, 0 \right) \)