Parabola of the Form y=x²+c: Finding points on the graph of a function

To determine the points of intersection of the function $y = x^2 - 49$ with the x-axis, we set $y = 0$. This gives us the equation:

$x^2 - 49 = 0$

We can solve this equation by factoring or using the square root method:

• Recognize $x^2 - 49$ as a difference of squares since $49 = 7^2$.
• The equation can be rewritten and factored as: $(x - 7)(x + 7) = 0$.

Setting each factor equal to zero gives:

$x - 7 = 0$ or $x + 7 = 0$

This simplifies to:

$x = 7$ or $x = -7$

Thus, the points of intersection are where the function crosses the x-axis, at the coordinates $(7, 0)$ and $(-7, 0)$.

Referring to the given answer choices, the correct choice is:

$(7, 0), (-7, 0)$

Therefore, the points of intersection of the function $y = x^2 - 49$ with the x-axis are $(7, 0)$ and $(-7, 0)$.

To solve this problem, we need to determine where the given function intersects the y-axis. This occurs when the x-coordinate is 0. Let's find the y-coordinate when $x = 0$.

The function in question is $y = 2x^2 + 10$.
Substitute $x = 0$ into the function:

$y = 2(0)^2 + 10 = 2 \cdot 0 + 10 = 10$.

Since substituting $x = 0$ yields $y = 10$, the point of intersection on the y-axis is $(0, 10)$.

Therefore, the point where the function intersects the y-axis is $(0, 10)$.

To solve this problem and determine where the function $x^2 + y = 3$ intersects the y-axis, we follow these steps:

• Step 1: Since the y-axis is where the x-coordinate is zero, set $x = 0$ in the given equation.
• Step 2: Substitute $x = 0$ into the equation $x^2 + y = 3$, which simplifies to $0 + y = 3$.
• Step 3: From the simplified equation, solve for $y$. The result is $y = 3$.

This indicates that the point of intersection on the y-axis is $(0, 3)$.

Accordingly, the solution to the problem is that the graph intersects the y-axis at the point $(0, 3)$.

To determine where the function intersects the Y-axis, we will follow these steps:

• Step 1: Identify the given equation $y - 5x^2 = 20$.
• Step 2: Set $x = 0$ to find the Y-intercept.
• Step 3: Solve for $y$ when $x = 0$.

Let's apply these steps in detail:

Step 1: The equation provided is:

$y - 5x^2 = 20$

Step 2: Since the function intersects the Y-axis when $x = 0$, substitute $x = 0$ into the equation:

$y - 5(0)^2 = 20$

This simplifies to:

$y - 0 = 20$

Therefore, $y = 20$.

Step 3: The point where the function intersects the Y-axis is

$(0, 20)$.

Therefore, the function intersects the Y-axis at point $(0, 20)$.

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

• Step 1: Set $y = 0$ in the given equation $9 - y = 4x^2$ to get the simplified equation $9 = 4x^2$.
• Step 2: Rearrange the equation to isolate $x^2$, yielding $4x^2 = 9$.
• Step 3: Solve for $x^2$ by dividing each side by 4, resulting in $x^2 = \frac{9}{4}$.
• Step 4: Take the square root of both sides to find $x = \pm \frac{3}{2}$.

Therefore, the points of intersection with the x-axis are $(-\frac{3}{2}, 0)$ and $(\frac{3}{2}, 0)$.

Referring to the provided choices, this correctly corresponds to choice 3.

To determine the points of intersection of the function $y = 16 - x^2$ with the x-axis, we follow these steps:

• Step 1: Set the equation equal to zero since the x-axis corresponds to $y = 0$.
  $y = 16 - x^2 = 0$
• Step 2: Rearrange the equation to isolate $x^2$.
  Adding $x^2$ to both sides gives: $x^2 = 16$.
• Step 3: Solve the equation $x^2 = 16$ for $x$ by taking the square root of both sides.
  $x = \pm \sqrt{16}$ results in $x = \pm 4$.
• Step 4: Determine the intersection points. For $x = 4$, the point of intersection is $(4, 0)$, and for $x = -4$, the point of intersection is $(-4, 0)$.

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

This corresponds to the answer choice: $(-4,0),(4,0)$.

To solve for the intersection points of the function $y=-27+3x^2$ with the x-axis, follow these steps:

• Step 1: Set the function equal to zero because intersections with the x-axis occur when $y = 0$.
  So we solve $-27 + 3x^2 = 0$.
• Step 2: Simplify and solve the equation.
  Start with $3x^2 = 27$.
• Step 3: Divide both sides by 3 to isolate $x^2$.
  $x^2 = 9$.
• Step 4: Solve for $x$ by taking the square root of both sides.
  Thus, $x = \pm \sqrt{9}$, so $x = 3$ or $x = -3$.

Therefore, the points of intersection are $(3, 0)$ and $(-3, 0)$.

The correct choices from the answer list are 1: $(3,0)$ and 2: $(-3,0)$. Therefore, answer choice 4: Answers $a$ and $b$ are correct is correct.

To determine the intersection points of the function $y = x^2 + 4$ with the x-axis, we set the equation to zero, i.e., find where $x^2 + 4 = 0$.

Let's solve the equation:

• $x^2 + 4 = 0$
• Subtract 4 on both sides: $x^2 = -4$
• Since $x^2 = -4$ has no real solutions (the square of a real number cannot be negative), there are no real $x$-intercepts.

Therefore, the parabola defined by $y = x^2 + 4$ does not intersect the x-axis.

The solution to the problem is No solution.

To determine the point where the function $y + 10 = 3x^2$ intersects the y-axis, follow these steps:

• Step 1: Identify that the y-axis intersection occurs where $x = 0$.
• Step 2: Substitute $x = 0$ into the function equation.
• Step 3: Solve for $y$.

Now, let's go through these steps:

Step 1: Set $x = 0$.

Step 2: Substitute into the equation:
$y + 10 = 3(0)^2$, which simplifies to:
$y + 10 = 0$.

Step 3: Solve for $y$:
$y = -10$.

Therefore, the point of intersection with the y-axis is $(0, -10)$.

To find where the function intersects the y-axis, we substitute $x = 0$ into the equation $x^2 + 3 - y = 0$.

• Substitute $x = 0$ into the function: $(0)^2 + 3 - y = 0$.
• Simplify: $0 + 3 - y = 0$ becomes $3 - y = 0$.
• Solving for $y$, we get: $y = 3$.

This means the function intersects the y-axis at the point $(0, 3)$.

Therefore, the solution to the problem is $(0, 3)$, corresponding to choice 3.