Problem 10 - Beginning and Intermediate Algebra

Question

For quadratic function, identify the vertex, axis of symmetry, and \(x\)- and \(y\)-intercepts. Then, graph the function. \(h(x)=(x+2)^{2}+7\)

Step-by-Step Solution

Answer

The vertex of the quadratic function \(h(x)=(x+2)^2+7\) is \((-2,7)\). The axis of symmetry is \(x=-2\). There are no real \(x\)-intercepts, as the parabola does not intersect the \(x\)-axis. The \(y\)-intercept is at the point \((0,11)\). To graph the function, plot the vertex, axis of symmetry, and \(y\)-intercept, then sketch the parabola with a vertical axis and symmetry about the axis of symmetry line.

Step 1: Identify the Vertex

The given quadratic function is in vertex form, which is \(h(x) = a(x-h)^2+k\). Comparing the given function \(h(x)=(x+2)^2+7\) with the vertex form, we can identify the vertex as \((-2, 7)\), since \(h = -2\) and \(k = 7\).

Step 2: Identify the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two equal halves. For a function in the vertex form, the equation of the axis of symmetry is given as \(x = h\). Since we have already found the vertex \((-2, 7)\), the axis of symmetry can be represented as \(x = -2\).

Step 3: Identify the \(x\)-intercepts

The \(x\)-intercepts are the points where the function intersects the \(x\)-axis, which means the output, \(h(x)\), should be equal to 0. So, we need to solve the following equation for \(x\): \[(x+2)^2+7 = 0\] Subtract 7 from both sides: \[(x+2)^2 = -7\] Since the square of a number cannot be negative, there is no real solution for the \(x\)-intercepts. This means the parabola does not intersect the \(x\)-axis.

Step 4: Identify the \(y\)-intercept

The \(y\)-intercept is the point where the function intersects the \(y\)-axis. This means the input, \(x\), should be equal to 0. So, we need to find the value of \(h(x)\) when \(x = 0\): \[h(0) = (0 + 2)^2 + 7\] \[h(0) = (2)^2 + 7\] \[h(0) = 4 + 7\] \[h(0) = 11\] So, the \(y\)-intercept is at the point \((0,11)\).

Step 5: Graph the Function

Now, with all the information we've gathered, we can graph the given quadratic function \(h(x)=(x+2)^2+7\). On the coordinate plane: 1. Plot the vertex at point \((-2,7)\). 2. Draw the axis of symmetry as a dotted line at \(x=-2\). 3. Plot the \(y\)-intercept at point \((0,11)\). 4. Since there are no \(x\)-intercepts (real solutions), the parabola doesn't intersect the \(x\)-axis. 5. Sketch the parabola, making sure that it has a vertical axis and is symmetric about the axis of symmetry line. And now we have the correct graph of the given function with all the key information.

Key Concepts

Vertex Form

The vertex form of a quadratic function is expressed as \( f(x) = a(x - h)^2 + k \). This representation is incredibly useful because it allows you to directly identify the vertex of the parabola. The vertex is the point \((h, k)\), where the parabola either reaches its highest or lowest value, depending on the direction it opens.

• For the quadratic function \( h(x) = (x+2)^2 + 7 \), you can rewrite it as \( h(x) = (x - (-2))^2 + 7 \).
• From this, the vertex is found at \((-2, 7)\).

Knowing the vertex is essential since it also helps determine the axis of symmetry of the parabola.

Axis of Symmetry

The axis of symmetry is a vertical line that runs through the vertex of a parabola, perfectly splitting it into two mirror-image halves. It can be found using the formula \( x = h \), where \( h \) is the x-coordinate of the vertex. This line is a critical concept when analyzing parabolas and understanding their properties.

• For the function in our exercise, \( h(x) = (x+2)^2 + 7 \), the vertex is \((-2, 7)\).
• Therefore, the axis of symmetry is \( x = -2 \).

The axis of symmetry's role is crucial in graphing as it guides the shape and direction of the parabola.

Graphing Parabolas

Graphing a parabola involves using its vertex and axis of symmetry as key starting points. These components guide you in plotting additional points to shape the curve.

• For the equation \( h(x)= (x+2)^2+7 \), start by plotting the vertex \((-2, 7)\).
• Draw a dotted vertical line through \( x = -2 \) to illustrate the axis of symmetry.
• Identify additional points if needed, using symmetrical properties about the axis.

Remember that the direction in which the parabola opens (up or down) depends on the value of \( a \) in the quadratic function; in our case, \( a = 1 \), so the parabola opens upwards.

Intercepts in Quadratic Equations

Intercepts provide specific points where the graph intersects the x-axis and y-axis. Finding intercepts can greatly aid in fully sketching a parabola and understanding its behavior across the coordinate plane.

• The \( x \)-intercepts are found by setting the quadratic equation to zero and solving for \( x \). For \( h(x)=(x+2)^2+7 \), when set to zero, it results in a negative number under the square, indicating no real \( x \)-intercepts.
• The \( y \)-intercept is found by setting \( x = 0 \) in the function, yielding the point \( (0, 11) \).

These intercepts help confirm the shape and position of the parabola on a graph. Since our parabola doesn't have real \( x \)-intercepts, it underscores that the curve doesn't cross the x-axis.