# How do I find the y-intercept of a parabola?

Sep 22, 2014

If given a graph: the y-intercept passes through y-axis.

Example:
If a graph passes the y-axis at -1, then the y-intercept is -1.

If you can not see the y-axis, you need to chose two points and create the equation of the parabola in standard or vertex form (using vertex and another point).

Example on how to create an equation from the graph of the parabola (when y-axis cannot be seen):
Vertex: (3,4)
Point: (5, -4)
-4 = $a {\left(5 - 3\right)}^{2}$+4
-4 = $a {\left(2\right)}^{2}$ + 4
-4 = $a \left(4\right)$ + 4
-4 = 4a + 4
-4 - 4 = 4a
-8 = 4a
$- \frac{8}{4}$ = $\frac{4 a}{4}$
-2 = a
Equation: y = $- 2 {\left(x - 3\right)}^{2}$+4 (refer below to find the y-intercept)

If an equation: When a point is on the y-axis, the x value is 0. Therefore it you want to plug in 0 for your x values. y = $a {x}^{2}$ + bx + c, has a y-intercept at c.

Example (vertex Form y = $a {\left(x - h\right)}^{2}$+k ):
y = $3 {\left(x - 4\right)}^{2}$ + 5
y = $3 {\left(0 - 4\right)}^{2}$ + 5
y = $3 {\left(- 4\right)}^{2}$ + 5
y = 3(16) +5
y = 48 + 5
y = 53
Y-intercept is 53

Example (Standard Form y = $a {x}^{2}$ + bx + c):
y = $2 {x}^{2}$ -3x + 4
y = $2 {\left(0\right)}^{2}$ -3(0) + 4
y = 2(0) - 3(0) + 4
y = 0 - 0 + 4
y = 4 (the c value)
Y-intercept is 4