Given:
y=x^2-10x+17 is a quadratic equation in standard form:
ax^2-10x+17,
where:
a=1, b=-10, and c=17
Vertex: the maximum or minimum point of the parabola
axis of symmetry and x value for vertex: x=(-b)/(2a)
x=(-(-10))/(2*1)
x=10/2
x=5
To find the y value of the vertex, substitute 5 for x into the equation and solve for y.
y=1xx5^2-10(5)+17
y=25-50+17
y=-8
The vertex is (5,-8). Since a>0, the vertex is the minimum point of the parabola, and the parabola opens upward.
X-Intercepts: the value of x when y=0
To find the x-intercepts, substitute 0 for y and solve for x using the quadratic formula.
0=x^2-10x+17
x=(-b+-sqrt(b^2-4ac))/(2a)
Plug in the known values.
x=(-(-10)+-sqrt(-10^2-4*1*17))/(2*1)
Simplify.
x=(10+-sqrt(100-68))/2
x=(10+-sqrt32)/2
Prime factorize 32.
x=(10+-sqrt(color(red)2xxcolor(red)2xxcolor(blue)2xxcolor(blue)2xx2))/2
Simplify.
x=(10+-color(red)2xxcolor(blue)2sqrt2)/2
x=(10+-4sqrt2)/2
Simplify.
x=(color(red)cancel(color(black)(10))^5+-color(red)cancel(color(black)(4))^2sqrt2)/color(red)cancel(color(black)(2))^1
x=5+-2sqrt2
x=5+2sqrt2,5-2sqrt2
x-intercepts: (5+2sqrt2,0),(5-2sqrt2,0)
Y-Intercept: the value of y when x=0
To find the y-intercept, substitute 0 for x and solve for y.
y=1xx0^2-10(0)+17
y=0+0+17
y=17
y-intercept is (0,17)
Summary
axis of symmetry: 5
vertex: (5,-8)
x-intercepts: (5+2sqrt2,0),(5-2sqrt2,0)
y-intercept: (0,17)
graph{y=x^2-10x+17 [-15.68, 16.34, 5.38, 21.4]}
