Parabola of the Form y=(x-p)²+k: Finding a stationary point

To find the intersection of the function $y = (x+5)^2 + 2$ with the y-axis, we follow these steps:

• Step 1: Identify that the intersection with the y-axis occurs when $x = 0$.
• Step 2: Substitute $x = 0$ into the equation $y = (x+5)^2 + 2$.
• Step 3: Perform the substitution and simplify: $y = (0+5)^2 + 2$.
• Step 4: Simplify further: $y = 5^2 + 2 = 25 + 2 = 27$.

Thus, the intersection point with the y-axis is $(0, 27)$.

The correct answer is option 4: $(0, 27)$.

The problem asks us to find where the parabola given by $y = (x-3)^2 - 4$ intersects the y-axis. The intersection with the y-axis occurs where $x = 0$. Let's find the value of $y$ by substituting $x = 0$ into the equation:

$y = (0 - 3)^2 - 4$

Simplify inside the parentheses:

$y = (-3)^2 - 4$

Calculate $(-3)^2$:

$y = 9 - 4$

Subtract 4 from 9:

$y = 5$

Thus, the intersection of the function with the y-axis occurs at the point $(0, 5)$.

The correct answer from the choices provided is $(0,5)$.

To solve this problem, we need to determine where the parabola $y = (x+2)^2 - 4$ intersects the x-axis. This occurs where $y = 0$.

Step 1: Set the equation equal to zero to find the x-intercepts:
$0 = (x+2)^2 - 4$

Step 2: Simplify the equation:
$(x+2)^2 = 4$

Step 3: Solve for $x$ by taking the square root of both sides:
$x+2 = \pm 2$

Step 4: Solve each equation for $x$:
1. $x+2 = 2$ leads to $x = 0$
2. $x+2 = -2$ leads to $x = -4$

Therefore, the points of intersection are $(-4, 0)$ and $(0, 0)$, where the parabola intersects the x-axis.

The correct answer to the problem is $(-4, 0), (0, 0)$.

To determine where the parabola intersects the x-axis, we solve for x when $y = 0$ in the function $y = (x+3)^2 + 4$.

Set $y = 0$:

$0 = (x+3)^2 + 4$

Subtract 4 from both sides to isolate the square term:

$(x+3)^2 = -4$

We now assess the equation $(x+3)^2 = -4$. A square is always non-negative, meaning that no real number squared gives a negative result. Hence, there's no real value of x satisfying this equation.

Thus, the parabola has no intersection points with the x-axis.

Therefore, the correct answer is that there is no intersection.

To solve the problem of finding the intersection of the function $y = (x+2)^2 + 3$ with the Y-axis, we will follow these steps:

Let's begin with identifying the point where the function intersects the Y-axis. A point on the Y-axis always has $x = 0$. So, we’ll substitute this value into the function to find the corresponding y-coordinate.

Substitute $x = 0$ into the function:

$y = (0+2)^2 + 3$

First, calculate $(0+2)^2$:

$(0+2)^2 = 2^2 = 4$

Then add 3 to the result:

$y = 4 + 3 = 7$

Thus, the intersection point on the Y-axis is $(0, 7)$.

Therefore, the solution to the problem is $(0, 7)$.

To solve this problem, we'll follow these steps:

• Step 1: Identify the equation of the function $y = (x+4)^2 - 1$.

• Step 2: Since we're finding the intersection with the Y-axis, set $x = 0$.

• Step 3: Substitute $x = 0$ into the equation and solve for $y$.

Now, let's work through each step:

Step 1: The equation of the function is already given as $y = (x+4)^2 - 1$.

Step 2: To find the Y-intercept, let $x = 0$.

Step 3: Substitute $x = 0$ into the equation:

$\begin{aligned} y &= (0 + 4)^2 - 1 \\ &= 4^2 - 1 \\ &= 16 - 1 \\ &= 15 \end{aligned}$

Therefore, the intersection with the Y-axis is at $(0, 15)$.

To solve this problem, follow these steps:

The problem asks to find the intersection of the parabola with the x-axis given by the equation $y = (x+1)^2 - 5$. The x-intercepts occur where $y = 0$.

Let's solve the equation for $y = 0$:

$(x+1)^2 - 5 = 0$

Simplify and solve for $x$:

• Add 5 to both sides: $(x+1)^2 = 5$
• Take the square root of both sides: $x+1 = \pm \sqrt{5}$
• Solve for $x$ by subtracting 1 from both sides: $x = -1 \pm \sqrt{5}$

This gives two solutions:

• $x = -1 + \sqrt{5}$
• $x = -1 - \sqrt{5}$

Therefore, the x-intercepts are $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$.

The correct answer is then the pair $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$, which matches choice 3:

$(-\sqrt{5}-1,0),(\sqrt{5}-1,0)$

Therefore, the solution to the problem is $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$

To solve the problem of finding the intersection of the function $y = (x + 3)^2 - 9$ with the x-axis, we need to set $y = 0$ because the x-axis is defined by $y = 0$.

Starting with the equation:

$(x + 3)^2 - 9 = 0$

Our goal is to solve for $x$. Let's simplify the equation:

Add 9 to both sides:

$(x + 3)^2 = 9$

Next, take the square root of both sides:

$x + 3 = \pm 3$

This gives us two equations to solve:

• $x + 3 = 3$
• $x + 3 = -3$

Solving these equations, we find:

For $x + 3 = 3$:

$x = 0$

For $x + 3 = -3$:

$x = -6$

Thus, the x-intercepts are $(0, 0)$ and $(-6, 0)$. These are the points where the graph intersects the x-axis.

The correct answer, matching the choices given, is choice 3: $(0, 0), (-6, 0)$.

To find the intersection of the function $y = (x+5)^2 - 49$ with the x-axis, we need to solve the equation for $y = 0$.

Set the equation equal to zero:

$(x+5)^2 - 49 = 0$

Add 49 to both sides:

$(x+5)^2 = 49$

Take the square root of both sides, remembering to consider both the positive and negative roots:

$x + 5 = 7$ or $x + 5 = -7$

Solve for $x$ in both cases:

• For $x + 5 = 7$:
  $x = 7 - 5$
  $x = 2$
• For $x + 5 = -7$:
  $x = -7 - 5$
  $x = -12$

Therefore, the x-intercepts of the function are $(-12, 0)$ and $(2, 0)$.

Thus, the function intersects the x-axis at these points.

$(-12, 0), (2, 0)$

To find the intersection of the function $y = (x - 7)^2 - 40$ with the $y$-axis, we set $x = 0$ since this is the defining property of the $y$-axis.

Substitute $x = 0$ into the function:

• Compute $y = (0 - 7)^2 - 40$.
• Calculate $y = (-7)^2 - 40$.
• This simplifies to $y = 49 - 40$.
• Thus, $y = 9$.

The point of intersection on the $y$-axis is therefore $(0, 9)$.

Therefore, the solution to the problem is $(0, 9)$.