Zeros of Quadratic Functions Practice Problems & Solutions

To solve for the x-intercepts of the function $f(x) = -x^2 + 5x + 6$, we will find the roots of the quadratic equation $-x^2 + 5x + 6 = 0$.

Let's attempt to factor this quadratic equation first. Rewrite the equation as follows:

$-x^2 + 5x + 6 = 0$.

To factor, we look for two numbers that multiply to $-6$ (the product of $a$ and $c$, where $a = -1$ and $c = 6$) and add to $5$ (the middle coefficient $b$).

The numbers that satisfy this condition are $-1$ and $6$.

Thus, the quadratic can be factored as:

$(x - 6)(x + 1) = 0$.

Setting each factor equal to zero gives us:

$x - 6 = 0$ or $x + 1 = 0$.

Solving these equations, we find:

$x = 6$ and $x = -1$.

Thus, the points A and B, the x-intercepts of the function, are:

$(-1, 0)$ and $(6, 0)$.

Therefore, the solution to the problem is $(-1, 0), (6, 0)$.

To solve for the points A and B, we need to find the roots of the function $f(x) = x^2 - 6x + 5$ where $f(x) = 0$.

Let's proceed step-by-step:

• Step 1: Set the function to zero
  We begin by setting the equation to zero: $x^2 - 6x + 5 = 0$.
• Step 2: Factor the quadratic
  We need to factor the expression. We look for two numbers that multiply to $c = 5$ and add to $b = -6$. These numbers are $-1$ and $-5$.
• Step 3: Write the factorization
  Therefore, we can write the quadratic as: $(x - 1)(x - 5) = 0$.
• Step 4: Solve for the roots
  Set each factor equal to zero: \begin{align*} x - 1 &= 0 \\ x &= 1 \end{align*} \begin{align*} x - 5 &= 0 \\ x &= 5 \end{align*} The roots are $x = 1$ and $x = 5$.
• Step 5: Identify the Points A and B
  The points A and B, where the function intersects the x-axis, are $(1, 0)$ and $(5, 0)$.

Thus, the coordinates of points A and B are $(1,0),(5,0)$, which matches choice 1.

Let's solve the problem by following the outlined analysis:

• Step 1: Identify important points on the parabola.
• Step 2: Calculate the y-intercept by evaluating $f(0)$.
• Step 3: Confirm understanding of the vertex form and its characteristics.

Step 1: Identify the significant points on the function.
The function given is $f(x) = x^2 - 8x + 16$.

This function can be seen as
$f(x) = (x - 4)^2$.

This format not only indicates it is always non-negative but also reveals the vertex is located at $x = 4$, importantly, with $f(x) = 0$.

Step 2: Calculate the y-intercept.
Evaluate the function at $x = 0$:

$f(0) = 0^2 - 8 \cdot 0 + 16 = 16$.
So, the y-intercept is $(0, 16)$.

Thus, point A, which is often labeled at a crucial intercept, corresponds to the y-intercept of the function. The calculation confirms that point A is $(0, 16)$.

Therefore, the solution to the problem is $(0,16)$.

To solve the exercise, first note that point C lies on the X-axis.

Therefore, to find it, we need to understand what is the X value when Y equals 0.

Let's set the equation equal to 0:

0=x²-8x+16

We'll use the preferred method (trinomial or quadratic formula) to find the X values, and we'll discover that

X=4

To solve for points A and B, we find the x-intercepts of the function by setting:

$f(x) = x^2 - 3x - 4 = 0$

We check if it can be factored:

Factor $x^2 - 3x - 4$. The factors of -4 that add to -3 are -4 and 1.

Thus, factor the function as $(x - 4)(x + 1) = 0$.

Set each factor to zero:

• $x - 4 = 0$ implies $x = 4$
• $x + 1 = 0$ implies $x = -1$

These are the x-intercepts, or roots, of the quadratic function.

Therefore, the coordinates of points A and B, where the function intersects the x-axis, are $A(-1, 0)$ and $B(4, 0)$.

The correct choice corresponding to these points is option 3: $A(-1,0), B(4,0)$.

Thus, the solution to the problem is $A(-1,0), B(4,0)$.