Find the x and y Intercepts y=-6x^2-5x-1
This problem is asking for the points where the graph of the given quadratic equation intersects the x-axis and y-axis. The x-intercepts (also called 'roots') are the values of x where the equation y = -6x^2 - 5x - 1 equals zero, while the y-intercept is the value of y when x is zero. Because the equation is quadratic, you may expect two x-intercepts (unless the discriminant is zero, in which case there would be a single x-intercept, or it is negative, in which case there would be no real x-intercepts). The y-intercept is straightforward to find, as it requires substituting x with zero and solving for y.
$y = - 6 x^{2} - 5 x - 1$
Answer
Solution:
Step:1
Determine the x-intercepts.
Step:1.1
Set $y$ to $0$ and solve for $x$: $0 = -6x^2 - 5x - 1$.
Step:1.2
Proceed to solve the quadratic equation.
Step:1.2.1
Rewrite the equation as $-6x^2 - 5x - 1 = 0$.
Step:1.2.2
Begin factoring the quadratic expression.
Step:1.2.2.1
Extract $-1$ from each term: $-(6x^2 + 5x + 1) = 0$.
Step:1.2.2.2
Attempt to factor by grouping.
Step:1.2.2.2.1
Split the middle term to factor by grouping, ensuring the two terms multiply to $6 \cdot 1 = 6$ and add up to $5$.
Step:1.2.2.2.2
Factor out the common factor from each binomial: $-(2x(3x + 1) + 1(3x + 1)) = 0$.
Step:1.2.2.2.3
Factor out the common binomial factor: $-(3x + 1)(2x + 1) = 0$.
Step:1.2.3
Set each factor equal to $0$: $3x + 1 = 0$ and $2x + 1 = 0$.
Step:1.2.4
Solve $3x + 1 = 0$ for $x$: $x = -\frac{1}{3}$.
Step:1.2.5
Solve $2x + 1 = 0$ for $x$: $x = -\frac{1}{2}$.
Step:1.2.6
The solutions are $x = -\frac{1}{3}$ and $x = -\frac{1}{2}$.
Step:1.3
Express the x-intercepts as ordered pairs: $(-\frac{1}{3}, 0)$ and $(-\frac{1}{2}, 0)$.
Step:2
Identify the y-intercepts.
Step:2.1
Set $x$ to $0$ and solve for $y$: $y = -6(0)^2 - 5 \cdot 0 - 1$.
Step:2.2
Simplify the equation to find $y$: $y = -1$.
Step:2.3
Express the y-intercept as an ordered pair: $(0, -1)$.
Step:3
Combine the intercepts.
List the x-intercepts and y-intercept: $(-\frac{1}{3}, 0)$, $(-\frac{1}{2}, 0)$, and $(0, -1)$.
Knowledge Notes:
To find the x-intercepts of a function, we set $y$ to $0$ and solve the resulting equation for $x$. The x-intercepts are the points where the graph of the function crosses the x-axis.
To find the y-intercepts, we set $x$ to $0$ and solve for $y$. The y-intercepts are the points where the graph of the function crosses the y-axis.
The quadratic equation $ax^2 + bx + c = 0$ can be solved by factoring, completing the square, or using the quadratic formula. In this case, we attempted to factor the equation.
Factoring by grouping is a method used when a polynomial does not factor easily. It involves rearranging and grouping terms to find common factors.
The distributive property, $a(b + c) = ab + ac$, is used in the factoring process.
The solutions to the quadratic equation are the values of $x$ that make the equation true, and these correspond to the x-intercepts of the graph.
Ordered pairs are used to represent points on a coordinate plane, with the x-coordinate first and the y-coordinate second, in the form $(x, y)$.
