I can help you with graphing quadratics in intercept form. Here's a step-by-step guide:
Step 1: Identify the intercepts
The intercepts are the points where the parabola crosses the x-axis and y-axis. In intercept form, the x-intercepts are given by the factors of the quadratic expression. For example, in the equation f(x) = 2(x+4)(x+6), the x-intercepts are -4 and -6. To find the y-intercept, set x = 0 and solve for y.
Step 2: Find the vertex
The vertex is the point where the parabola reaches its maximum or minimum value. The x-coordinate of the vertex is the average of the x-intercepts. In the equation f(x) = 2(x+4)(x+6), the x-intercepts are -4 and -6, so the x-coordinate of the vertex is (-4 + -6)/2 = -5. To find the y-coordinate of the vertex, substitute the x-coordinate into the equation and solve for y.
Step 3: Plot the intercepts and vertex
Mark the intercepts and vertex on the coordinate plane.
Step 4: Sketch the parabola
Draw a smooth curve that passes through the intercepts and vertex. The parabola should be symmetric about the vertical line passing through the vertex.
Example:
Let's graph the equation f(x) = 2(x+4)(x+6).
Step 1: Identify the intercepts
The x-intercepts are -4 and -6. To find the y-intercept, set x = 0 and solve for y:
f(0) = 2(0+4)(0+6) = 48
So the y-intercept is (0, 48).
Step 2: Find the vertex
The x-coordinate of the vertex is (-4 + -6)/2 = -5. To find the y-coordinate of the vertex, substitute x = -5 into the equation:
f(-5) = 2(-5+4)(-5+6) = 2
So the vertex is (-5, 2).
Step 3: Plot the intercepts and vertex
Plot the intercepts (-4, 0), (-6, 0), and (0, 48), and the vertex (-5, 2) on the coordinate plane.
Step 4: Sketch the parabola
Draw a smooth curve that passes through the intercepts and vertex. The parabola should be symmetric about the vertical line passing through the vertex.
The graph of the equation f(x) = 2(x+4)(x+6) is a parabola that opens upwards and has intercepts at (-4, 0), (-6, 0), and (0, 48). The vertex of the parabola is at (-5, 2).