Question

NEED ANSWER ASAP AND MARKING BRANLIEST

Answer 1

The given quadratic equation forms a parabola when graphed, a U-shaped line.

To find the minimum value of y, we could find the y-coordinate of the vertex.

What is the vertex?

The vertex represents the maximum or minimum point on a parabola. It tells us the optimal y-value of the function and the x-value at which it occurs.

For the given quadratic, $y=x^2+6x+7$ the optimal y-value, or the y-coordinate of the vertex, will be a minimum. The $x^2$ value is positive, meaning the parabola will open upwards. (Please see below for an image representation).

Different Ways to Approach the Problem

There are multiple ways to approach this question.

One way is to rearrange the equation, which is currently in standard form, to vertex form.

• Standard form for a quadratic: $y=ax^2+bx+c$

• Vertex form for a quadratic: $y=a(x-h)^2+k$ where (h,k) is the vertex

Another way is to factor the equation to first find the x-intercepts of the graph, find the x-coordinate of the vertex, then at the end, plug the x-coordinate into the equation to find the minimum y-value.

The second strategy is a bit redundant for this type of question, so we can go about solving it by using the first method.

Finding the Vertex of this Parabola

$y=x^2+6x+7$

First, we can group $x^2$ and $6x$ by using a set of parentheses:

$y=(x^2+6x)+7$

Then, we can complete the square:

$y=(x^2+6x+(\dfrac{6}{2})^2)^2+7-(\dfrac{6}{2})^2$

$y=(x^2+6x+9)^2+7-9\\\\y=(x^2+6x+9)^2-2$

Apply the rule $a^2+2ab+b^2=(a+b)^2$:

$y=(x+3)^2-2$

Therefore, the vertex of the quadratic is (-3,-2).

This makes the minimum y-value -2. The x-value at which this occurs is -3.

Answer:

The minimum y-value -2. The x-value at which this occurs is -3.