Determine the zeros of the quadratic function y=x^2 + 15x + 36 . (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)

Question

smaller zero: x= ___
; greater zero: x= ___

Answer 1

To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we need to solve the equation \( x^2 + 15x + 36 = 0 \).

We can use the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 15 \), and \( c = 36 \).

First, calculate the discriminant: \[ b^2 - 4ac = 15^2 - 4(1)(36) = 225 - 144 = 81 \]

Now, we can plug the values into the quadratic formula: \[ x = \frac{-15 \pm \sqrt{81}}{2(1)} = \frac{-15 \pm 9}{2} \]

This gives us two solutions:

1. For the positive case: \[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]

2. For the negative case: \[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]

Thus, the zeros of the function are:

• Smaller zero: \( x = -12 \)
• Greater zero: \( x = -3 \)

Final answer:

• smaller zero: \( x = -12 \)
• greater zero: \( x = -3 \)

Answer 2

Determine the zeros of the quadratic function y=−x^2 + 16x − 39
. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)
smaller zero: x= ___
; greater zero: x= ___

Answer 3

To determine the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we need to solve the equation:

\[ -x^2 + 16x - 39 = 0 \]

We can rearrange this to:

\[ x^2 - 16x + 39 = 0 \]

Now, we will apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -16 \), and \( c = 39 \).

First, we calculate the discriminant: \[ b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100 \]

Now we can use the quadratic formula: \[ x = \frac{-(-16) \pm \sqrt{100}}{2(1)} = \frac{16 \pm 10}{2} \]

This gives us two solutions:

1. For the positive case: \[ x = \frac{16 + 10}{2} = \frac{26}{2} = 13 \]

2. For the negative case: \[ x = \frac{16 - 10}{2} = \frac{6}{2} = 3 \]

Thus, the zeros of the function are:

• Smaller zero: \( x = 3 \)
• Greater zero: \( x = 13 \)

Final answer:

• smaller zero: \( x = 3 \)
• greater zero: \( x = 13 \)

Answer 4

Determine the vertex and axis of symmetry of y=x^2 − 10x + 3
.(1 point)
vertex: ( ___ ); axis of symmetry: x= ___

Answer 5

To find the vertex and axis of symmetry of the quadratic function \( y = x^2 - 10x + 3 \), we can use the vertex formula and the property of the axis of symmetry.

The vertex \( (h, k) \) of a quadratic in the form \( y = ax^2 + bx + c \) can be found using the formulas:

• The x-coordinate of the vertex \( h \) is given by: \[ h = -\frac{b}{2a} \]
• The y-coordinate of the vertex \( k \) can be found by substituting \( h \) back into the quadratic function.

In our case:

• \( a = 1 \)
• \( b = -10 \)
• \( c = 3 \)

1. Calculate \( h \): \[ h = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]

2. Calculate \( k \) by substituting \( h \) into the function: \[ k = (5)^2 - 10(5) + 3 = 25 - 50 + 3 = -22 \]

Thus, the vertex is \( (5, -22) \).

3. Find the axis of symmetry: The axis of symmetry is given by the line \( x = h \).

So in this case: \[ \text{axis of symmetry} : x = 5 \]

Final answer:

• vertex: \( (5, -22) \)
• axis of symmetry: \( x = 5 \)

Answer 6

What point is symmetrical to (−1,−3)
for the graph of y=x^2 + 12x + 8
?(1 point)
( ___ )

Answer 7

To find the point that is symmetrical to \((-1, -3)\) with respect to the graph of the quadratic function \(y = x^2 + 12x + 8\), we need to follow these steps:

1. Find the vertex of the quadratic: The vertex of a parabola in the form \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate of the vertex: \[ h = -\frac{b}{2a} \] In this case:

   • \(a = 1\)
   • \(b = 12\)

   Therefore, \[ h = -\frac{12}{2 \cdot 1} = -\frac{12}{2} = -6 \]

   To find the y-coordinate of the vertex \(k\), substitute \(h\) back into the equation: \[ k = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28 \]

   Thus, the vertex of the parabola is \((-6, -28)\).

2. Find the line of symmetry: The line of symmetry for the parabola is \(x = -6\).

3. Determine the symmetrical point: The point \((-1, -3)\) is on the left side of the line of symmetry \(x = -6\). To find its symmetrical point, we need to find the distance from \(-1\) to \(-6\): \[ \text{Distance} = -6 - (-1) = -6 + 1 = -5 \] The symmetrical point will be \(5\) units to the right of \(-6\): \[ x_{\text{symmetrical}} = -6 + 5 = -1 \]

   The y-coordinate remains the same since symmetry across the vertex leads to mirroring vertically at the vertex level, so we calculate the y-value of the original point and then reflect over the vertex.

4. Find the y-coordinate of the symmetrical point: Substitute \(x = -1\) into the original equation to find \(y\): \[ y = (-1)^2 + 12(-1) + 8 = 1 - 12 + 8 = -3 \]

Thus, the symmetrical point of \((-1, -3)\) across the vertex (\(-6, -28\)) is at the same vertical height as the original:

Therefore, the symmetrical point to \((-1, -3)\) is: \[ (-11, -3) \]

So, the answer is: ( \( -11, -3 \) )

Answer 8

What are the key points on the graph of y=x2−16x+48
? Name the vertex, x-intercepts, and y-intercept.(2 points)
vertex: ( ___ )

(smaller) x-intercept: ( ___ )

(larger) x-intercept: ( ___ )

y-intercept: ( ___ )

Answer 9

To find the key points on the graph of the quadratic function \( y = x^2 - 16x + 48 \), we will identify the vertex, the x-intercepts, and the y-intercept.

Vertex

The vertex of a quadratic function in the form \( y = ax^2 + bx + c\) can be found with the following formulas:

1. Calculate the x-coordinate of the vertex \( h \): \[ h = -\frac{b}{2a} \] In this case:

   • \( a = 1 \)
   • \( b = -16 \)

   Therefore: \[ h = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]

2. Calculate the y-coordinate \( k \) by substituting \( h \) back into the function: \[ k = 8^2 - 16(8) + 48 = 64 - 128 + 48 = -16 \]

   Thus, the vertex is: \[ \text{vertex: } (8, -16) \]

X-Intercepts

The x-intercepts occur when \( y = 0 \): \[ x^2 - 16x + 48 = 0 \]

We can solve this using factoring: \[ (x - 12)(x - 4) = 0 \]

Setting each factor to zero gives us: \[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

So the x-intercepts are:

• Smaller x-intercept: \( (4, 0) \)
• Larger x-intercept: \( (12, 0) \)

Y-Intercept

The y-intercept occurs when \( x = 0 \): \[ y = 0^2 - 16(0) + 48 = 48 \]

So the y-intercept is: \[ \text{y-intercept: } (0, 48) \]

Summary of Key Points

• Vertex: \( (8, -16) \)
• Smaller x-intercept: \( (4, 0) \)
• Larger x-intercept: \( (12, 0) \)
• Y-intercept: \( (0, 48) \)

Final Answer:

• vertex: \( (8, -16) \)
• (smaller) x-intercept: \( (4, 0) \)
• (larger) x-intercept: \( (12, 0) \)
• y-intercept: \( (0, 48) \)