Respuesta :
[tex]\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
\begin{array}{lcccl}
y = & 2x^2& +8x& -3\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad
\left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( -\cfrac{8}{2(2)}~~,~~-3-\cfrac{8^2}{4(2)} \right)\implies \left( \cfrac{-8}{4}~~,~~-3-\cfrac{64}{8} \right)
\\\\\\
\left(-2~~,~~-3-8 \right)\implies (-2~~,~~-11)[/tex]
Answer:
The vertex is (-2, -11 ).
Step-by-step explanation:
Here, the given equation is,
[tex]y = 2x^2 + 8x - 3[/tex] -----(1)
Which is a parabola along x-axis,
Since, the standard form of parabola along x-axis is,
[tex]y=a(x-h)^2 + k[/tex]
Where (h,k) is the vertex of the parabola,
By solving it,
[tex]y = ax^2 - 2ahx + ah^2 + k[/tex] -----(2),
By comparing equation (1) and (2),
We get,
a = 2, -2ah = 8 and ah² + k = - 3
⇒ -2(2) h = 8 ⇒ - 4 h = 8 ⇒ h = -2
And, (2)(-2)² + k = - 3
⇒ 8 + k = - 3
⇒ k = -11
Hence, the vertex of the given equation is (-2, -11 )
