Quadratic Function Practice Problems - Parabola Vertex & Graph

Here we have a quadratic equation.

A quadratic equation is always constructed like this:

$y = ax²+bx+c$

Where a, b, and c are generally already known to us, and the X and Y points need to be discovered.

Firstly, it seems that in this formula we do not have the C,

Therefore, we understand it is equal to 0.

$c = 0$

a is the coefficient of X², here it does not have a coefficient, therefore

$a = 1$

$b= 10$

is the number that comes before the X that is not squared.

To solve this problem, we will follow these steps:

• Step 1: Identify each term in the given function $y = 2x^2 + 3$.
• Step 2: Compare the equation to the standard quadratic form $y = ax^2 + bx + c$.
• Step 3: Determine the coefficients $a$, $b$, and $c$.
• Step 4: Match these coefficients to the correct multiple-choice option.

Step 1: The given function is $y = 2x^2 + 3$. There is no $x$ term present.

Step 2: Compare this with the standard form $y = ax^2 + bx + c$:

• The coefficient of $x^2$ is $a = 2$.
• The coefficient of $x$ is $b = 0$ because there is no $x$ term.
• The constant term is $c = 3$.

Step 3: Therefore, the coefficients are $a = 2$, $b = 0$, and $c = 3$.

Step 4: Review the multiple-choice options provided:

• Choice 1: $a = 0$, $b = 2$, $c = 3$
• Choice 2: $a = 0$, $b = 3$, $c = 2$
• Choice 3: $a = 2$, $b = 0$, $c = 3$
• Choice 4: $a = 3$, $b = 0$, $c = 2$

The correct choice is Choice 3: $a = 2$, $b = 0$, $c = 3$.

Therefore, the solution to the problem is the values $a = 2$, $b = 0$, $c = 3$ which correspond to choice 3.

To solve this problem, we'll clearly delineate the given expression and compare it to the standard quadratic form:

• Step 1: Recognize the standard form of a quadratic equation as $y = ax^2 + bx + c$.
• Step 2: Compare the given equation $y = x^2 - 6x + 4$ to the standard form.
• Step 3: Identify coefficients:
  - The coefficient of $x^2$ is $a = 1$.
  - The coefficient of $x$ is $b = -6$.
  - The constant term is $c = 4$.

Therefore, the coefficients for the quadratic function $y = x^2 - 6x + 4$ are $a = 1$, $b = -6$, and $c = 4$.

Among the provided choices, choice 3: $a=1,b=-6,c=4$ is the correct one.

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

• Identify the given quadratic function.
• Match it with the standard form of a quadratic equation $y = ax^2 + bx + c$.
• Extract the values of $a$, $b$, and $c$ directly from the comparison.

Now, let's work through each step:
Step 1: The given quadratic function is $y = 2x^2 - 3x - 6$.
Step 2: The standard form of a quadratic equation is $y = ax^2 + bx + c$.
Step 3: By matching the given quadratic function with the standard form:

- The coefficient of $x^2$ is $2$, so $a = 2$.
- The coefficient of $x$ is $-3$, so $b = -3$.
- The constant term is $-6$, so $c = -6$.

Therefore, the solution to the problem is $a = 2$, $b = -3$, $c = -6$.

To solve this problem, we will identify values of $a$, $b$, and $c$ in the quadratic function:

• Step 1: Note the given equation $y = 3x^2 - 81$.
• Step 2: Compare it to the standard quadratic form $y = ax^2 + bx + c$.
• Step 3: Match coefficients to find $a$, $b$, and $c$.

Now, let's work through each step:
Step 1: The given equation is $y = 3x^2 - 81$.
Step 2: Compare this to the standard form, $y = ax^2 + bx + c$. In this equation:
- The coefficient of $x^2$ is 3, hence $a = 3$.
- There is no $x$ term, which means $b = 0$.
- The constant term is $-81$, hence $c = -81$.

Therefore, the solution to the problem is $a = 3, b = 0, c = -81$.