Question

multiple choice 2.5 points select the graph that represents the function: $x^2 + 6x - 27$

Explanation:

Step1: Analyze the function type

The function \( y = x^2 + 6x - 27 \) is a quadratic function. For a quadratic function in the form \( y = ax^2+bx + c \), if \( a>0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards. Here, \( a = 1>0 \), so the parabola should open upwards. But the given graph in the image opens downwards (since it has a maximum point), so this graph is not the correct one. Wait, maybe there are other graphs not shown? Wait, the problem says "select the graph", but maybe the user missed showing other options. Wait, but let's check the vertex and intercepts.

First, find the vertex of the parabola. The x - coordinate of the vertex of \( y=ax^2 + bx + c \) is \( x=-\frac{b}{2a} \). For \( y=x^2 + 6x - 27 \), \( a = 1 \), \( b = 6 \), so \( x=-\frac{6}{2\times1}=- 3 \). Then the y - coordinate is \( y=(-3)^2+6\times(-3)-27=9 - 18 - 27=-36 \). So the vertex is at \( (-3,-36) \).

Find the x - intercepts by setting \( y = 0 \): \( x^2+6x - 27=0 \). Factor the quadratic: \( x^2+6x - 27=(x + 9)(x - 3)=0 \). So the x - intercepts are \( x=-9 \) and \( x = 3 \). The y - intercept is when \( x = 0 \), \( y=-27 \).

Now, the graph shown in the image has a maximum (opens down, so \( a<0 \)), but our function has \( a = 1>0 \) (opens up). So the correct graph should open upwards, have vertex at \( (-3,-36) \), x - intercepts at \( x=-9 \) and \( x = 3 \), and y - intercept at \( y=-27 \).

But since the given graph in the image opens down, it's incorrect. Maybe the other options (not shown) have a parabola opening upwards with the correct vertex and intercepts.

Wait, maybe the user made a mistake in the image. But assuming that there are other graphs, the correct graph should be a parabola opening upwards, with vertex at \( (-3,-36) \), crossing the x - axis at \( x=-9 \) and \( x = 3 \), and y - axis at \( y=-27 \).

Step2: Compare with the given graph

The given graph in the image has a maximum (opens down), so it's for a quadratic with \( a<0 \). Our function has \( a = 1>0 \), so the correct graph is not the one shown. But since the problem is about selecting the graph, and maybe the other options (not shown) have the correct upward - opening parabola.

Wait, maybe the original function was miswritten? If the function was \( y=-x^2 + 6x - 27 \), then \( a=-1<0 \), opens down, vertex at \( x =-\frac{6}{2\times(-1)} = 3 \), \( y=-9 + 18-27=-18 \), which doesn't match the given graph's vertex.

Alternatively, maybe the user's image is incomplete. But based on the analysis, the correct graph should open upwards, have vertex at \( (-3,-36) \), x - intercepts at \( x=-9 \) and \( x = 3 \), and y - intercept at \( y=-27 \).

Answer:

Since the given graph in the image opens downwards (incorrect for \( y = x^2+6x - 27 \) as it should open upwards), the correct graph (not shown here) should be a parabola opening upwards with vertex at \((-3, - 36)\), x - intercepts at \(x=-9\) and \(x = 3\), and y - intercept at \(y=-27\). If we assume there are other options, the correct graph will have these features.