Question

determine the x-intercepts

the x - intercepts are
(type integers or fractions. use a comma to separate answers as needed.)
choose the correct graph below. each graph is shown in a -10,10,1 by -10,10,1 viewing rectangle.

1. graph the parabola whose equation is given.

graph the parabola using the tool to the right.

Explanation:

For the equation \( y = x^2 - 5x + 4 \) to find x - intercepts:

Step 1: Set \( y = 0 \)

We know that at x - intercepts, the value of \( y = 0 \). So we set up the equation \( 0=x^{2}-5x + 4 \) or \( x^{2}-5x + 4=0 \)

Step 2: Factor the quadratic equation

We need to find two numbers that multiply to \( 4 \) and add up to \( - 5 \). The numbers are \( -1 \) and \( -4 \). So we can factor the quadratic as \( x^{2}-5x + 4=(x - 1)(x - 4) \)

Step 3: Solve for \( x \)

Using the zero - product property, if \( (x - 1)(x - 4)=0 \), then either \( x-1 = 0 \) or \( x - 4=0 \)

• If \( x-1=0 \), then \( x = 1 \)
• If \( x - 4=0 \), then \( x=4 \)

Step 1: Find the vertex

The x - coordinate of the vertex of a parabola given by \( y=ax^{2}+bx + c \) is \( x=-\frac{b}{2a} \). For \( y=x^{2}+8x + 15 \), \( a = 1 \), \( b = 8 \) and \( c=15 \)
\( x=-\frac{8}{2\times1}=-4 \)
To find the y - coordinate of the vertex, substitute \( x=-4 \) into the equation:
\( y=(-4)^{2}+8\times(-4)+15=16-32 + 15=-1 \)
So the vertex is at \( (-4,-1) \)

Step 2: Find the x - intercepts

Set \( y = 0 \), so \( x^{2}+8x + 15=0 \)
Factor the quadratic: We need two numbers that multiply to \( 15 \) and add up to \( 8 \). The numbers are \( 3 \) and \( 5 \). So \( x^{2}+8x + 15=(x + 3)(x+5) \)
Using the zero - product property, if \( (x + 3)(x + 5)=0 \), then \( x=-3 \) or \( x=-5 \)

Step 3: Find the y - intercept

Set \( x = 0 \), then \( y=0^{2}+8\times0+15 = 15 \), so the y - intercept is at \( (0,15) \)

To graph the parabola:

• Plot the vertex \( (-4,-1) \)
• Plot the x - intercepts \( (-5,0) \) and \( (-3,0) \)
• Plot the y - intercept \( (0,15) \)
• Since \( a = 1>0 \), the parabola opens upwards. Then draw a smooth curve passing through these points.

(For the multiple - choice graph part related to \( y=x^{2}-5x + 4 \)):
The parabola \( y=x^{2}-5x + 4 \) has x - intercepts at \( x = 1 \) and \( x = 4 \), and the vertex of \( y=x^{2}-5x + 4 \) is at \( x=\frac{5}{2}=2.5 \), \( y=(2.5)^{2}-5\times(2.5)+4=6.25-12.5 + 4=-2.25 \). The parabola opens upwards (because \( a = 1>0 \)). So we look for a parabola that opens upwards, has x - intercepts at \( 1 \) and \( 4 \) and vertex at \( (2.5,-2.25) \)

Answer:

The x - intercepts are \( 1,4 \)

For the equation \( y=x^{2}+8x + 15 \) to graph the parabola: