Question

Please help me with my math

Answer 1

`\textsf{$\boxed{\checkmark}\;\;y$-value\;of\;vertex\;is\;$-1$}`

`\textsf{$\boxed{\checkmark}$\;\;Minimum\;value\;occurs\;at\;$y = -1$}`

Explanation:

Given equation:

`y=x^2+8x+15`

As the given equation is quadratic with a positive leading coefficient, it is a parabola that opens upwards. Therefore, its vertex is its minimum point. This means that the minimum value of the range is the y-value of the vertex.

The x-value of the vertex of a parabola in the form y = ax² + bx + c is x = -b/2a. Therefore, the x-value of the vertex of the given equation is:

`\implies x=(-8)/(2(1))=-4`

To find the y-value of the vertex, substitute x = -4 into the equation:

`\begin{aligned}\implies y&=(-4)^2+8(-4)+15\\&=16-32+15\\&=-16+15\\&=-1\end{aligned}`

Therefore, the minimum y-value of the function is y = -1, so the range is y ≥ -1.

Therefore, the following are true statement about the given equation:

• y-value of vertex is -1
• Minimum value occurs at y = -1