Question

question 27 (mandatory) (1 point) ✓ saved determine which coordinate is the vertex of $f(x) = 4x^2 - 8x + 11$ without graphing the parabola. \\(\bigcirc\\) a) \\((1, 10)\\) \\(\bigcirc\\) b) \\((1, 12)\\) \\(\bigcirc\\) c) \\((1, 7)\\) \\(\bigcirc\\) d) \\((-1, 7)\\)

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \).
In the function \( f(x)=4x^{2}-8x + 11 \), we have \( a = 4 \), \( b=-8 \), and \( c = 11 \).
Substitute \( a = 4 \) and \( b=-8 \) into the formula for the x - coordinate of the vertex:
\( x=-\frac{-8}{2\times4}=\frac{8}{8} = 1 \)

Step2: Find the y - coordinate of the vertex

Now that we know the x - coordinate of the vertex is \( x = 1 \), we substitute \( x = 1 \) into the function \( f(x)=4x^{2}-8x + 11 \) to find the y - coordinate.
\( f(1)=4\times(1)^{2}-8\times(1)+11 \)
\(=4\times1-8 + 11\)
\(=4-8 + 11\)
\(=-4 + 11\)
\(=7\)

So the vertex of the parabola \( f(x)=4x^{2}-8x + 11 \) is \( (1,7) \).

Answer:

c) (1, 7)