Question

graph this function: $f(x) = x^2 - 4x - 5$
$x = 2$
find the $y$-coordinate of the vertex:
$f(2) = 2^2 - 4(2) - 5$.
$f(2) = -9$
step 3: plot the vertex.
step 4: factor the function to find the $x$-intercepts.
the $x$-intercepts are: $(5, 0)$ and $(-1, 0)$
step 5: plot the $x$-intercepts.
correct!

Explanation:

Step 1: Recall the vertex formula for a 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 \( f(x)=x^{2}-4x - 5 \), \( a = 1 \), \( b=-4 \). So \( x=-\frac{-4}{2\times1}=2 \).

Step 2: Calculate the y - coordinate of the vertex

Substitute \( x = 2 \) into the function \( f(x)=x^{2}-4x - 5 \). So \( f(2)=2^{2}-4\times2 - 5 \). First, calculate the powers and multiplications: \( 2^{2}=4 \), \( 4\times2 = 8 \). Then \( f(2)=4-8 - 5 \). Then perform the subtractions: \( 4-8=-4 \), \( -4 - 5=-9 \).

Step 3: Factor the quadratic function to find x - intercepts

We factor \( f(x)=x^{2}-4x - 5 \). We need two numbers that multiply to \( - 5 \) and add up to \( -4 \). The numbers are \( -5 \) and \( 1 \). So \( f(x)=(x - 5)(x+1) \). To find the x - intercepts, set \( f(x)=0 \), so \( (x - 5)(x + 1)=0 \). Then \( x-5 = 0 \) or \( x + 1=0 \), which gives \( x = 5 \) or \( x=-1 \). So the x - intercepts are \( (5,0) \) and \( (-1,0) \).

Answer:

The vertex is at \( (2,-9) \), and the x - intercepts are at \( (5,0) \) and \( (-1,0) \). To graph the function, plot the vertex \( (2,-9) \) and the x - intercepts \( (5,0) \) and \( (-1,0) \), then draw a parabola opening upwards (since \( a = 1>0 \)) passing through these points.