Question

graph this function: $f(x) = x^2 - 4x - 5$
step 2: find the vertex using the formula $x = \frac{-b}{2a}$.
$x = \frac{4}{2(1)}$
$x = 2$
find the $y$-coordinate of the vertex:
$f(2) = 2^2 - 4(2) - 5$
$f(2) = -9$
step 3: plot the vertex.

Explanation:

Step1: Identify the coefficients

For the quadratic function \( f(x) = ax^2 + bx + c \), here \( a = 1 \), \( b = -4 \), \( c = -5 \).

Step2: Calculate the x - coordinate of the vertex

Using the formula \( x = -\frac{b}{2a} \), substitute \( a = 1 \) and \( b = -4 \):
\( x = -\frac{-4}{2\times1}=\frac{4}{2}=2 \)

Step3: Calculate the y - coordinate of the vertex

Substitute \( x = 2 \) into the function \( f(x)=x^{2}-4x - 5 \):
\( f(2)=2^{2}-4\times2 - 5=4 - 8 - 5=-9 \)

Step4: Determine other points (optional, for graphing)

We can find more points by choosing different \( x \) values. For example, when \( x = 0 \):
\( f(0)=0^{2}-4\times0 - 5=-5 \)
When \( x = 1 \):
\( f(1)=1^{2}-4\times1 - 5=1 - 4 - 5=-8 \)
When \( x = 3 \):
\( f(3)=3^{2}-4\times3 - 5=9 - 12 - 5=-8 \)
When \( x = 4 \):
\( f(4)=4^{2}-4\times4 - 5=16 - 16 - 5=-5 \)

Step5: Plot the points

Plot the vertex \((2,-9)\) and the other points \((0,-5)\), \((1,-8)\), \((3,-8)\), \((4,-5)\) on the coordinate plane and draw a parabola through them.

Answer:

The vertex of the parabola \( f(x)=x^{2}-4x - 5 \) is \((2,-9)\), and the graph is a parabola opening upwards (since \( a = 1>0 \)) passing through points like \((0,-5)\), \((1,-8)\), \((3,-8)\), \((4,-5)\) etc.