Find the solutions to the equation f(x)=0. For the axis of symmetry, find the midpoint between the solutions. To find the y-intercept, find f(0).
Equation: f(x)=(x+4)(x-3) Graph:
Practice makes perfect
We are given the quadratic function f(x)=x^2+x-12 and we need to write it in intercept form.
f(x) = a(x-p)(x-q)
In the equation above, p and q represent the solutions to the equation f(x)=0. Hence, we begin by finding the solutions using the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2aIn our case, a= 1, b= 1 and c= -12. Let's substitute these values into the formula above.
The x-intercepts of the function are x= -1± 72. Let's separate them using the positive and negative signs.
x=-1 ± 7/2
x=-1 + 7/2
x=-1 - 7/2
x=6/2
x=-8/2
x=3
x=-4
The parabola intercepts the x-axis at -4 and 3. In other words, p=-4 and q=3. We are ready to rewrite the given function in intercept form.
f(x) = (x-(-4))(x-3) = (x+4)(x-3)
Next, we find the axis of symmetry by finding the midpoint between the x-intercepts.
Axis of Sym. ⟶ x = -4+3/2 = -1/2
From the above, we also know that the x-coordinate of the vertex is - 12 and the y-coordinate is equal to f(- 12). Let's find the y-coordinate.
We now know that the vertex is at (- 12,- 494). Finally, we can find the y-intercept by setting x=0 in f(x)=x^2+x-12.
y=f(0)=0^2+0-12 = -12
The graph intercepts the y-axis at (0,-12). With all this information, we can graph the function.
