Question

If \((a,-5)\) is a point on the graph of \(y=x^{2}+6 x,\) what is \(a ?\)

Short answer

a = -1; a = -5

Step-by-step solution

Identify the given point and equation

The given point is \(a, -5\), and the equation is \(y = x^2 + 6x\). We need to find the value of \(a\).

Substitute the y-coordinate into the equation

Substitute the \(-5\) from the point \(a, -5\) into the equation for \(y\): \(-5 = a^2 + 6a\).

Set up the quadratic equation

Rewrite the equation in standard form: \(a^2 + 6a + 5 = 0\).

Factor the quadratic equation

Factor the quadratic equation: \((a + 1)(a + 5) = 0\).

Solve for a

Set each factor equal to zero: \a + 1 = 0\ or \a + 5 = 0\. Solving these gives \(a = -1\) or \(a = -5\).

Key concepts

graph of a function

When it comes to understanding the graph of a function, think of it as a visual representation of a mathematical relationship. For instance, the function given in the exercise is a quadratic function: \(y = x^2 + 6x\).To graph this function, you plot points that satisfy the equation. In this case, the graph will be a parabola, a U-shaped curve that can open upwards or downwards based on the coefficient of the highest degree term (in our case, \(x^2\)) being positive (opening upwards) or negative (opening downwards). To better interpret the graph:

• Choose several x-values and calculate the corresponding y-values to get points.
• Plot these points on a coordinate plane.
• Connect the points with a smooth curve.

For our specific problem, knowing that \((a, -5)\) lies on this graph is crucial to find the unknown value \(a\).

substitute values

Substituting values is a key step in solving equations. It's like filling in the blanks. In this exercise, the given point is \((a, -5)\). We know the y-value of the point, so we substitute -5 into the equation \(y = x^2 + 6x\). Here we replace \(y\) with -5:

\( -5 = a^2 + 6a \).
This substitution allows us to form a new equation that we can solve to determine the value of \(a\).

factoring quadratics

Factoring quadratics is an essential technique in algebra. It's a method to express a quadratic equation in a product of two binomials. In our problem, after substituting the values into the equation, we have:

\(a^2 + 6a + 5 = 0\).
To factor this:

• Look for two numbers that multiply to +5 (the constant term) and add to +6 (the coefficient of the middle term after substituting and re-arranging).

The two numbers are +1 and +5, meaning we can write:

\((a + 1)(a + 5) = 0\).
Factoring helps simplify the problem and allows us to solve for the unknown values of \(a\). In this case, it splits the quadratic into simpler linear factors.

solving quadratic equations

Solving quadratic equations can often be broken down after factoring. From the factored form \((a + 1)(a + 5) = 0\), we use the Zero Product Property, which states if the product of two factors is zero, at least one of the factors must be zero.

• Set each factor equal to zero: \(a + 1 = 0\) or \(a + 5 = 0\).

Solve these simple equations to find the possible values of \(a\):

• \(a + 1 = 0\) gives \(a = -1\).
• \(a + 5 = 0\) gives \(a = -5\).

Hence, our values for \(a\) can be either \(-1\) or \(-5\).This straightforward process ensures we correctly solve any quadratic equation by methodically working through substitution, factoring, and solving.