Question

graph the equation $y = -x^2 + 10x - 24$ on the accompanying set of axes. you must plot 5 points including the roots and the vertex.
click to plot points. click points to delete them.
answer attempt 1 out of 2
you must answer all questions above in order to submit.

Explanation:

Step1: Find the roots (x-intercepts)

To find the roots, set \( y = 0 \):

$$ 0=-x^{2}+10x - 24 $$

Multiply both sides by -1:

$$ x^{2}-10x + 24=0 $$

Factor the quadratic:

$$ (x - 4)(x - 6)=0 $$

So, \( x=4 \) or \( x = 6 \). The roots are \( (4,0) \) and \( (6,0) \).

Step2: Find the vertex

The x - coordinate of the vertex of a quadratic \( y=ax^{2}+bx + c \) is \( x=-\frac{b}{2a} \). For \( y=-x^{2}+10x - 24 \), \( a=-1 \), \( b = 10 \).

$$ x=-\frac{10}{2\times(-1)}=\frac{-10}{-2}=5 $$

Substitute \( x = 5 \) into the equation to find the y - coordinate:

$$ y=-(5)^{2}+10\times5-24=-25 + 50-24=1 $$

The vertex is \( (5,1) \).

Step3: Find two more points

Let's find the y - intercept by setting \( x = 0 \):

$$ y=-(0)^{2}+10\times0-24=-24 $$

So, the point is \( (0,-24) \).

Let's find another point, say when \( x = 1 \):

$$ y=-(1)^{2}+10\times1-24=-1 + 10-24=-15 $$

The point is \( (1,-15) \). (We can also use \( x = 2 \), \( x = 3 \), etc. as long as we have 5 points. Here we use \( (0,-24) \) and \( (1,-15) \) along with the roots \( (4,0) \), \( (6,0) \) and vertex \( (5,1) \))

Now we can plot the points \( (0,-24) \), \( (1,-15) \), \( (4,0) \), \( (5,1) \), \( (6,0) \) and draw the parabola.

Answer:

To graph \( y=-x^{2}+10x - 24 \), plot the points: \( (0,-24) \), \( (1,-15) \), \( (4,0) \), \( (5,1) \), \( (6,0) \) and draw the parabola through them. (The key points are the roots \( (4,0) \), \( (6,0) \), vertex \( (5,1) \), and two additional points like \( (0,-24) \) and \( (1,-15) \))